g)\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right).\dfrac{7}{3}\)
   \(=\left(-\dfrac{3}{4}+-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right).\dfrac{7}{3}\)
\(=\left(-1+1\right).\dfrac{7}{3}=0.\dfrac{7}{3}=0\)

f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)

\(=\dfrac{3^3.5^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)

\(=\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)

\(=\dfrac{5^3}{6^3}\)

\(=\dfrac{125}{216}\)

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

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

\(=-1-1-1+4\)

=1

a) Ta có: =−2590+6490−8190=−2590+6490−8190

(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)

=−53+4229=−53+4229

1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1

1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)

\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)

\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)

\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)

=1

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)

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

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

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

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)

=\(\left(4-\dfrac{5}{12}\right).\dfrac{1}{2}+\dfrac{5}{24}\)

=\(4.\dfrac{1}{2}-\dfrac{5}{12}.\dfrac{1}{2}+\dfrac{5}{24}\)

=\(2-\dfrac{5}{24}+\dfrac{5}{24}=2\)

Giải:

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\) 

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

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

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

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\) 

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

` a/`

` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `

`b/`

`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `

`c/`

` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `

`= 4/3 :  ( 2 + 7/6 : 17/6 ) `

`= 4/3 : ( 2 + 7/6 xx 6/17 )`

`= 4/3 : ( 2 + 7/17 ) `

`= 4/3 : ( 34/17 + 7/17 ) `

`= 4/3 : 41/17 `

`= 4/3 xx 17/41 `

`= 68/123`

` d/` 

` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`

1/ ĐKXĐ : \(x\ne1\)

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

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)

Vậy...

b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)

\(\Leftrightarrow12-28x=1+x\)

\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)

Vậy....

c/ ĐKXĐ : \(x\ne0\)

\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2x^2-12=2x^2+3x\)

\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)

Vậy...

4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)

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

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)

\(\Leftrightarrow6x^2+4x-3x-2=5\)

\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)

Vậy....

5,6 Tương tự nhé !

1)ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow21x-9-2x+2=0\)

\(\Leftrightarrow19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)

Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)

2) ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)

\(\Leftrightarrow4\left(3-7x\right)=x+1\)

\(\Leftrightarrow12-28x-x-1=0\)

\(\Leftrightarrow-29x+11=0\)

\(\Leftrightarrow-29x=-11\)

\(\Leftrightarrow x=\dfrac{11}{29}\)

Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)

3) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-12=2x^2+6x\)

\(\Leftrightarrow2x^2-12-2x^2-6x=0\)

\(\Leftrightarrow-6x-12=0\)

\(\Leftrightarrow-6x=12\)

\(\Leftrightarrow x=-2\)

Vậy: S={-2}

a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Ai giúp vs