ko qui đồng nha

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 2: 

a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ

mà số nguyên tố chẵn duy nhất là 2

nên số lẻ còn lại là 599(thỏa ĐK)

Vậy: Hai số nguyên tố cần tìm là 2 và 599

b,Gọi ƯCLN(21n+4,14n+3)=d

21n+4⋮d ⇒42n+8⋮d

14n+3⋮d ⇒42n+9⋮d

(42n+9)-(42n+8)⋮d

1⋮d ⇒ƯCLN(21n+4,14n+3)=1

Vậy phân số 21n+4/14n+3 là phân số tối giản

2. Tính:

a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)

=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)

=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)

=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)

b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)

=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)

=\(\dfrac{13}{4}\)

3. Tìm x

a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)

\(\left(x-5\right).\left(x-5\right)=8.18\)

\(\left(x-5\right)^2=144\)

\(x-5=\sqrt{144}\)

\(x-5=12\)

\(x=12+5\)

\(x=17\)

b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)

\(x^{10}-2^{10}=x^8-2^8\)

\(x^{10}+x^8=2^{10}+2^8\)

\(\Rightarrow x=2\)

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

Lời giải:

a, Đặt \(A=\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(\Rightarrow A=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+...+\dfrac{-1}{9.10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+...-\dfrac{1}{9}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-3}{20}\)

\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{9.10}\\ =\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{10}\\=\dfrac{-1}{4}+\dfrac{1}{10}\\=\dfrac{-5}{20}+\dfrac{2}{20}\\=\dfrac{-3}{20}\)

Lời giải:

a)

\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+....+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+...+\frac{-1}{9.10}\)

\(=\frac{4-5}{4.5}+\frac{5-6}{5.6}+\frac{6-7}{6.7}+....+\frac{9-10}{9.10}\)

\(=\frac{1}{5}-\frac{1}{4}+\frac{1}{6}-\frac{1}{5}+\frac{1}{7}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{9}\)

\(=\frac{1}{10}-\frac{1}{4}=-\frac{3}{20}\)

b)

\(2B=5+\frac{8}{11}+\frac{3}{11}+\frac{1}{15}+\frac{13}{15.2}\)

\(=5+\frac{11-3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{15-2}{15.2}\)

\(=5+1-\frac{3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{1}{2}-\frac{1}{15}\)

\(=5+1+\frac{1}{2}=\frac{13}{2}\Rightarrow B=\frac{13}{4}\)

Thanks

mk làm câu a cách khác ; nhưng cũng = \(\dfrac{-3}{20}\)

a. Ta có:

b. Theo kết quả câu a,ta có:

a. Ta có:

b. Theo kết quả câu a,ta có:

1/

a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)

⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)

⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)

b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993

2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993

2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993

2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993

2.(1 − 1/x+1) = 3984/1993

1 − 1/x + 1= 3984/1993 :2

1 − 1/x+1 = 1992/1993

1/x+1 = 1 − 1992/1993

1/x+1=1/1993

<=>x+1 = 1993

<=>x+1=1993

<=> x+1=1993

<=> x = 1993-1

<=> x = 1992