\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)

\(\Leftrightarrow7-6x=3x-5\)

\(\Leftrightarrow7+5=3x+6x\)

\(\Leftrightarrow12=9x\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

Đề bài yêu cầu gì bạn?

A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

=> 3(x +3) < 5(5 -x)

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)

<=> 5(x - 4) + 2x = 2(x +1)

<=> 5x - 20 + 2x = 2x + 2

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)

\(\frac{3x-1}{2}-\frac{2-6x}{5}=\frac{1}{2}+\left(3x-1\right)\)

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

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

\(\Leftrightarrow\frac{-1}{10}\left(3x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow3x-1=-5\)

\(\Leftrightarrow3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy nghiệm duy nhất của phương trình là\(x=\frac{-4}{3}\)

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

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5\left(x+1\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+6-5\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+1\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}-\frac{\left(x+1\right)\left(6x+1\right)}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+1-\frac{1}{3}-\frac{6x+1}{6}\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy nghiệm duy nhất của phương trình là\(x=-1\)

\(\frac{2}{x-1}+\frac{5}{x+2}=\frac{13}{x^2+x-2}.\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{5\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow\frac{2x+4}{x^2+x-2}+\frac{5x-5}{x^2+x-2}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow\frac{7x-1}{x^2+x-2}=\frac{13}{x^2+x-2}\)

\(\Leftrightarrow7x-1=13\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

\(\frac{2x-1}{x-3}=\frac{6x-1}{3x+2}\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=\left(x-3\right)\left(6x-1\right)\)

\(\Leftrightarrow6x^2+4x-3x-2=6x^2-x-18x+3\)

\(\Leftrightarrow4x-3x+x+18x=3+2\)

\(\Leftrightarrow20x=5\)

\(\Leftrightarrow x=\frac{1}{4}\)

\(\left|-3x\right|=x-10\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=x-10\\-3x=10-x\end{cases}\Leftrightarrow\orbr{\begin{cases}-3x-x=-10\\-3x+x=10\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\-2x=10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-5\end{cases}}}}}\)

\(\left|3x-1\right|-x+2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=x-2\\3x-1=2-x\end{cases}\Leftrightarrow\orbr{\begin{cases}3x-x=-2+1\\3x+x=2+1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-1\\4x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{4}\end{cases}}}}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+5=3x+x^2+1\\x^2-3x+5=-3x-x^2-1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(-3x-3x\right)=1-5\\\left(x^2+x^2\right)=-1-5\end{cases}\Leftrightarrow\orbr{\begin{cases}-6x=-4\\2x^2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=-3\text{(vô lí)}\end{cases}}}}}\)