Ta có: \(\left|97\dfrac{2}{3}-123\dfrac{3}{5}+97\dfrac{2}{5}-125\dfrac{1}{3}\right|\)

\(=\left|97\left(\dfrac{2}{3}+\dfrac{2}{5}\right)-125\cdot\left(\dfrac{3}{5}+\dfrac{1}{3}\right)\right|\)

\(=\left|194\cdot\dfrac{8}{15}-125\cdot\dfrac{14}{15}\right|\)

\(=\left|\dfrac{-66}{5}\right|=\dfrac{66}{5}\)

a) S = 1.2 + 2.3 + 3.4 + ... + 99.100

S có thể được viết lại thành:

S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)

= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98

= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)

Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:

S = n(n+1)(2n+1)/6

Với n = 99, ta có:

S = 99.100.199/6 = 331650

Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:

S = n(n+1)/2

Với n = 98, ta có:

S = 98.99/2 = 4851

Do đó, S = 331650 - 4851 = 326799

b) B = 4924.12517.28−530.749.45529.162.748

B có thể được viết lại thành:

B = (4924.12517.28) / (530.749.45529.162.748)

B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)

B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529

B = 108 / 45529

c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101

C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101

C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)

C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)

C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)

d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018

D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)

<=> \(25x+10-80x+10=24x+12-30\)

<=> \(25x-80x-24x=12-30-10-10\)

<=> \(-79x=-38\)

<=> \(x=\dfrac{-38}{-79}\)

\(x=\dfrac{38}{79}\)

b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)

<=> \(30x-12x+30+5x+40=210+10x-10\)

<=> \(30x-12x+5x-10x=210-10-30-40\)

<=> \(13x=130\)

<=> \(x=\dfrac{130}{13}\)

\(x=10\)

c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)

<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)

<=> \(28x+28+60x+120+105x+420+2520=0\)

<=> \(28x+60x+105x=-28-120-420-2520\)

<=> \(193x=-3088\)

<=> \(x=\dfrac{-3088}{193}\)

\(x=-16\)

d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)

<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)

<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)

<=> \(22968x=8199576\)

<=> \(x=\dfrac{8199576}{22968}\)

\(x=357\)

Đề là giải PT nha các bn

a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)

b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)

c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)

a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) =  \(\dfrac{6}{7}\)  

b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)          

c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)

A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0

A = 0*

*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)

Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)

hay A=50

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

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

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)

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

\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)

\(=\dfrac{10}{12}\)

\(=\dfrac{5}{6}\)

\(---\)

b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)

\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)

\(=18+\dfrac{3015}{2}\)

\(=\dfrac{36}{2}+\dfrac{3015}{2}\)

\(=\dfrac{3051}{2}\)

\(---\)

c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)

\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(---\)

d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)

\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)

\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)

\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)

\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)

\(=-\dfrac{25}{41}\)

#\(Toru\)

cảm ơn nhiều nha vừa kịp giờ lun