Bài 3:

\(A=75\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\)

Đặt: \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\\ 4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\\ 3B=4^{2005}-1\\ B=\dfrac{4^{2005}-1}{3}\)

\(=>A=75\cdot\dfrac{4^{2005}-1}{3}+25\\ =25\left(4^{2005}-1\right)+25\\ =25\cdot\left(4^{2005}-1+1\right)\\ =25\cdot4^{2005}\\ =25\cdot4\cdot4^{2004}\\ =100\cdot4^{2004}\) 

=> A chia hết cho 100 

Bài 1:

a: \(\dfrac{5\cdot3^{11}+4\cdot3^{17}}{3^9\cdot5^2-3^9\cdot2^3}=\dfrac{3^{11}\cdot\left(5+4\cdot3^6\right)}{3^9\left(5^2-2^3\right)}\)

\(=3^2\cdot\dfrac{5+4\cdot729}{25-8}=3^2\cdot\dfrac{2921}{17}=\dfrac{26289}{17}\)

b: \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{47\cdot50}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{47\cdot50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{1}{3}\cdot\dfrac{24}{50}=\dfrac{24}{150}=\dfrac{8}{50}=\dfrac{4}{25}\)

c: \(1+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=\dfrac{2}{2}+\dfrac{2}{6}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\right)\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)

d: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{99^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-98}{99}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

Bài 8:

\(4)-\dfrac{20}{7}:\dfrac{5}{21}\\ =-\dfrac{20}{7}\cdot\dfrac{21}{5}\\ =-4\cdot3\\ =-12\\ 5)-\dfrac{8}{5}:\dfrac{-12}{7}\\ =\dfrac{-8}{5}\cdot\dfrac{-7}{12}\\ =\dfrac{14}{15}\\ 6)\dfrac{-12}{21}:\dfrac{1}{6}\\ =\dfrac{-4}{7}\cdot6\\ =-\dfrac{24}{7}\)

Bài 10:

1: \(4,5\cdot\left(-\dfrac{4}{9}\right)=-\dfrac{9}{2}\cdot\dfrac{4}{9}=-\dfrac{4}{2}=-2\)

2: \(2,4\cdot\left(-3\dfrac{4}{7}\right)=\dfrac{-12}{5}\cdot\dfrac{25}{7}=\dfrac{-12}{7}\cdot\dfrac{25}{5}=-5\cdot\dfrac{12}{7}=-\dfrac{60}{7}\)

3: \(0,2\cdot\dfrac{-15}{4}=\dfrac{1}{5}\cdot\dfrac{-15}{4}=-\dfrac{15}{5}\cdot\dfrac{1}{4}=-\dfrac{3}{4}\)

4: \(\left(-3,5\right):\left(-2\dfrac{4}{5}\right)=\dfrac{3.5}{2\dfrac{4}{5}}=\dfrac{3.5}{2.8}=\dfrac{5}{4}\)

5: \(\dfrac{-5}{23}:\left(-2\right)=\dfrac{5}{23}:2=\dfrac{5}{23\cdot2}=\dfrac{5}{46}\)

6: \(1,25:\left(-3\dfrac{1}{8}\right)=1.25:\dfrac{-25}{8}=1.25\cdot\dfrac{-8}{25}=-\dfrac{10}{25}=-\dfrac{2}{5}\)

Bài 9:

1: \(-3\dfrac{1}{9}\cdot\dfrac{4}{21}=\dfrac{-28}{9}\cdot\dfrac{4}{21}=-\dfrac{28}{21}\cdot\dfrac{4}{9}=-\dfrac{4}{9}\cdot\dfrac{4}{3}=-\dfrac{16}{27}\)

2: \(-\dfrac{3}{4}\cdot2\dfrac{1}{2}=-\dfrac{3}{4}\cdot\dfrac{5}{2}=\dfrac{-15}{8}\)

3: \(-\dfrac{8}{15}\cdot1\dfrac{1}{4}=-\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{-40}{60}=-\dfrac{2}{3}\)

4: \(-\dfrac{11}{15}:1\dfrac{1}{10}=-\dfrac{11}{15}:\dfrac{11}{10}=-\dfrac{11}{15}\cdot\dfrac{10}{11}=-\dfrac{10}{15}=-\dfrac{2}{3}\)

5: \(1\dfrac{1}{5}:\left(-2\dfrac{1}{5}\right)=\dfrac{6}{5}:\dfrac{-11}{5}=\dfrac{6}{5}\cdot\dfrac{5}{-11}=\dfrac{6}{-11}=-\dfrac{6}{11}\)

6: \(\left(-3\dfrac{1}{7}\right):\left(-1\dfrac{6}{49}\right)=\dfrac{-22}{7}:\dfrac{-55}{49}=\dfrac{22}{7}\cdot\dfrac{49}{55}\)

\(=\dfrac{22}{55}\cdot\dfrac{49}{7}=7\cdot\dfrac{2}{5}=\dfrac{14}{5}\)

Bài 8:

1: \(\dfrac{-20}{41}\cdot\dfrac{-4}{5}=\dfrac{20}{5}\cdot\dfrac{4}{41}=4\cdot\dfrac{4}{41}=\dfrac{16}{41}\)

2: \(\dfrac{-24}{5}\cdot\dfrac{15}{-8}=\dfrac{-24}{-8}\cdot\dfrac{15}{5}=3\cdot3=9\)

3: \(\dfrac{-4}{34}\cdot\dfrac{17}{-24}=\dfrac{4}{24}\cdot\dfrac{17}{34}=\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{12}\)

Bài 6:

1: \(2\dfrac{3}{5}-1\dfrac{2}{3}=\dfrac{13}{5}-\dfrac{5}{3}=\dfrac{39}{15}-\dfrac{25}{15}=\dfrac{14}{15}\)

2: \(3\dfrac{3}{7}+2\dfrac{1}{2}=\dfrac{24}{7}+\dfrac{5}{2}=\dfrac{48}{14}+\dfrac{35}{14}=\dfrac{83}{14}\)

3: \(-3\dfrac{1}{2}-2\dfrac{1}{4}=\dfrac{-7}{2}-\dfrac{9}{4}=\dfrac{-14}{4}-\dfrac{9}{4}=-\dfrac{23}{4}\)

4: \(-2\dfrac{1}{2}-3\dfrac{1}{4}=-\dfrac{5}{2}-\dfrac{13}{4}=-\dfrac{10}{4}-\dfrac{13}{4}=-\dfrac{23}{4}\)

5: \(-4\dfrac{1}{2}+2\dfrac{3}{10}=-\dfrac{9}{2}+\dfrac{23}{10}=-\dfrac{45}{10}+\dfrac{23}{10}=-\dfrac{22}{10}=-\dfrac{11}{5}\)

6: \(-6\dfrac{1}{7}-\left(-7\dfrac{1}{6}\right)=-6-\dfrac{1}{7}+7+\dfrac{1}{6}\)

\(=1-\dfrac{1}{7}+\dfrac{1}{6}=\dfrac{6}{7}+\dfrac{1}{6}=\dfrac{43}{42}\)

Bài 7:

\(1)\dfrac{2}{7}+\dfrac{6}{21}-\dfrac{3}{14}\\ =\dfrac{12}{42}+\dfrac{12}{42}-\dfrac{12}{42}\\ =\dfrac{12}{42}\\ =\dfrac{2}{7}\\ 2)-\dfrac{7}{2}+\dfrac{3}{4}-\dfrac{17}{12}\\ =\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}\\ =\dfrac{-50}{12}=\dfrac{-25}{6}\\ 3)\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{1}{12}+\dfrac{3}{12}+\dfrac{8}{12}=\dfrac{12}{12}=1\\ 4)\dfrac{1}{3}+\dfrac{-4}{5}-\dfrac{8}{15}\\ =\dfrac{5}{15}+\dfrac{-12}{15}-\dfrac{8}{15}=-\dfrac{15}{15}=-1\\ 5)\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{2}{6}\\ =\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{1}{3}\\ =\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{-3}{4}\\ =1+\dfrac{-3}{4}\\ =\dfrac{1}{4}\\ 6\text{ })-\dfrac{5}{18}+\dfrac{5}{45}-\dfrac{9}{6}\\ =-\dfrac{5}{18}+\dfrac{1}{9}-\dfrac{3}{2}\\ =-\dfrac{5}{18}+\dfrac{2}{18}-\dfrac{27}{18}\\ =-\dfrac{30}{18}\\ =-\dfrac{5}{3}\)

Bài 8:

1: \(\dfrac{-20}{41}\cdot\dfrac{-4}{5}=\dfrac{20}{5}\cdot\dfrac{4}{41}=4\cdot\dfrac{4}{41}=\dfrac{16}{41}\)

2: \(\dfrac{-24}{5}\cdot\dfrac{15}{-8}=\dfrac{-24}{-8}\cdot\dfrac{15}{5}=3\cdot3=9\)

3: \(\dfrac{-4}{34}\cdot\dfrac{17}{-24}=\dfrac{4}{24}\cdot\dfrac{17}{34}=\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{12}\)

Bài 6:

1: \(2\dfrac{3}{5}-1\dfrac{2}{3}=\dfrac{13}{5}-\dfrac{5}{3}=\dfrac{39}{15}-\dfrac{25}{15}=\dfrac{14}{15}\)

2: \(3\dfrac{3}{7}+2\dfrac{1}{2}=\dfrac{24}{7}+\dfrac{5}{2}=\dfrac{48}{14}+\dfrac{35}{14}=\dfrac{83}{14}\)

3: \(-3\dfrac{1}{2}-2\dfrac{1}{4}=\dfrac{-7}{2}-\dfrac{9}{4}=\dfrac{-14}{4}-\dfrac{9}{4}=-\dfrac{23}{4}\)

4: \(-2\dfrac{1}{2}-3\dfrac{1}{4}=-\dfrac{5}{2}-\dfrac{13}{4}=-\dfrac{10}{4}-\dfrac{13}{4}=-\dfrac{23}{4}\)

5: \(-4\dfrac{1}{2}+2\dfrac{3}{10}=-\dfrac{9}{2}+\dfrac{23}{10}=-\dfrac{45}{10}+\dfrac{23}{10}=-\dfrac{22}{10}=-\dfrac{11}{5}\)

6: \(-6\dfrac{1}{7}-\left(-7\dfrac{1}{6}\right)=-6-\dfrac{1}{7}+7+\dfrac{1}{6}\)

\(=1-\dfrac{1}{7}+\dfrac{1}{6}=\dfrac{6}{7}+\dfrac{1}{6}=\dfrac{43}{42}\)

Tớ sẽ làm mẫu cho cậu 1 số bài nhé:

a) \(A=\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{25-24}{24.25}\)

   \(A=\dfrac{6}{5.6}-\dfrac{5}{5.6}+\dfrac{7}{6.7}-\dfrac{6}{6.7}+...+\dfrac{25}{24.25}-\dfrac{24}{24.25}\)

   \(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

   \(A=\dfrac{1}{5}-\dfrac{1}{25}\)

   \(A=\dfrac{4}{25}\)

Bài 4; 2 và câu d bài 1 cậu sẽ cần phải đưa tử về hiệu giữa 2 thừa số ở mẫu.

\(\dfrac{4}{5}\) K = \(\dfrac{7-3}{3.7}+\dfrac{11-7}{7.11}+\dfrac{15-11}{11.15}=...+\dfrac{85-81}{81.85}+\dfrac{89-85}{85.89}\)

\(\dfrac{4}{5}K=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{81}-\dfrac{1}{85}+\dfrac{1}{85}-\dfrac{1}{89}\)

\(\dfrac{4}{5}K=\dfrac{1}{3}-\dfrac{1}{89}\)

\(\dfrac{4}{5}K=\dfrac{43}{147}\)

\(K=\dfrac{43}{147}\div\dfrac{4}{5}\)

\(K=\dfrac{215}{588}\)

Với bài 3 thì cậu chỉ cần đảo vị trí từ dưới lên trên là được nhé.

Bài 5: (Viết lại tổng E). Khoảng cách giữa 2 thừa số ở mẫu là 6, cậu hãy nhân tử với 6, tính sau đó : 6 nhé.

→ E = \(\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

Bài 6. Quan sát:

\(3\left(\dfrac{1}{5}-\dfrac{3}{x-2}\right)=\dfrac{24}{35}\) và tương tự như câu b, luôn là cái đầu tiên - cái cuối cùng.

Bài 7. Cậu trừ 1 ở cả 2 vế rồi nhân \(\dfrac{1}{2}\).

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\)

= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\)

Cậu cứ làm như những bài trên nhé.

Bài 1: 

\(a,A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=\dfrac{1}{5}-\dfrac{1}{25}=>\dfrac{5}{25}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

\(b,B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

       \(=1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

       \(=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

       \(=1.\left(1-\dfrac{1}{101}\right)\)

       \(=\dfrac{100}{101}\)

\(c,K=\dfrac{4}{11.16}+\dfrac{4}{16.21}+\dfrac{4}{21.26}+...+\dfrac{4}{61.66}\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11.16}+\dfrac{1}{16.21}+\dfrac{1}{21.26}+...+\dfrac{1}{61.66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{66}\right)\)

       \(=\dfrac{4}{5}.\dfrac{5}{66}=>4.\dfrac{1}{66}\)

       \(=\dfrac{4}{66}=\dfrac{2}{33}\)

\(d,N=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)

        \(=2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

        \(=2.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

        \(=2.\left(1-\dfrac{1}{101}\right)\)

        \(=2.\dfrac{100}{101}\)

        \(=\dfrac{200}{101}\)

Bài 2:

\(K=\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{81.85}+\dfrac{5}{85.89}\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{81.85}+\dfrac{1}{85.89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\dfrac{86}{267}\)

    \(=\dfrac{215}{534}\)

Bài 3:

\(A=\dfrac{1}{25.24}+\dfrac{1}{24.23}+...+\dfrac{1}{7.6}+\dfrac{1}{6.5}\)

    \(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{25}\)

    \(=\dfrac{4}{25}\)

Bài 4 :

\(A=\dfrac{5}{3.6}+\dfrac{5}{6.9}+\dfrac{5}{9.12}+...+\dfrac{5}{99.102}\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{99.102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\dfrac{11}{34}\)

    \(=\dfrac{55}{102}\)

Bài 5 :

Sửa đề :\(a,E=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

        \(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

        \(=\dfrac{1}{6}.\left(\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\dfrac{36}{37}\)

        \(=\dfrac{6}{37}\)

\(b,C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

       \(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{13}\)

       \(=\dfrac{10}{39}\)

Bài 6 :

\(a,\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\) 

    \(\dfrac{3}{2}\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)

    \(\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=-\dfrac{9}{35}\)

                  \(-9\left(x+2\right)=1.35\)

                  \(-9\left(x+2\right)=35\)

                         \(x+2=35:-9\)

                        \(x+2=\dfrac{-35}{9}\)

                        \(x\)        \(=\dfrac{-35}{9}-2\)

                        \(x\)        \(=\dfrac{-53}{9}\)

Vậy \(x=\dfrac{-53}{9}\)

\(b,\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{x+3}\right)\)                                        \(=\dfrac{1}{9}\)

   \(\dfrac{1}{6}-\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{6}-\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{18}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{2}{36}\)

      ⇒   \(3.\left(x+3\right)=36\)

                 \(x+3=36:3\)

                 \(x+3=12\) 

                 \(x\)       \(=12-3\)

                 \(x\)       \(=9\)

Vậy \(x=9\)

Bài 7:

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(=>\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

             \(1-\dfrac{1}{x+1}=\dfrac{3980}{1991}:2\)

             \(1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=\dfrac{1}{1991}\)

           \(=>x+1=1991\)

                  \(x\)       \(=1991-1\)

                  \(x\)       \(=1990\)

Vậy \(x=1990\)

Các bạn giúp mình nha gợi ý : lưu ý trong 3 số tự nhiên liên tiếp có 1số chia hết cho 3

\(x-\dfrac{4}{8}=\dfrac{5}{18}\\ x=\dfrac{5}{18}+\dfrac{4}{8}\\ x=\dfrac{5}{18}+\dfrac{1}{2}\\ x=\dfrac{5}{18}+\dfrac{9}{18}\\ x=\dfrac{5+9}{18}\\ x=\dfrac{14}{18}\\ x=\dfrac{7}{9}\)

Chữ số 6 có giá trị là 60

=>6 là chữ số hàng chục

5 không nằm ở hàng đơn vị

mà 5 không là chữ số hàng chục

nên 5 là chữ số hàng trăm

=>Chữ số hàng đơn vị là 7

Vậy: Số cần tìm là 567

a) Để x là số hữu tỉ dương thì:

\(\dfrac{13-n}{-5}>0\)
Mà: `-5<0` 

`=>13-n<0`

`=>n>13` 

b) Để x là số hữu tỉ âm thì:

`(13-n)/-5<0`

Mà:  `-5<0`

`=>13-n>0`

`=> n<13` 

c) Đê  x không phải số hữu tỉ âm cũng không phải số hữu tỉ dương thì:

\(x=0=>\dfrac{13-n}{-5}=0\\ =>13-n=0\\ =>n=13\)

Bài 2:

Để \(\dfrac{m+2}{5};\dfrac{m-5}{-6}\) đều là các số dương thì

\(\left\{{}\begin{matrix}\dfrac{m+2}{5}>0\\\dfrac{m-5}{-6}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+2>0\\m-5< 0\end{matrix}\right.\)

=>-2<m<5

mà m nguyên

nên \(m\in\left\{-1;0;1;2;3;4\right\}\)
 Bài 3:

Để \(\dfrac{1-m}{-13};\dfrac{5-m}{3}\) đều là các số âm thì

\(\left\{{}\begin{matrix}\dfrac{1-m}{-13}< 0\\\dfrac{5-m}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m-1}{13}< 0\\\dfrac{m-5}{3}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m-1< 0\\m-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 1\\m>5\end{matrix}\right.\)

=>\(m\in\varnothing\)