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 =\(\frac{22}{5}\)

\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)

<=> \(\frac{4}{\left(x-1\right)\left(x-2\right)}-\frac{3}{2x^2-6x+1}+1=0\)

<=> 4(2x² - 6x + 1) - 3(x - 1)(x - 2) + (x - 1)(x - 2)(2x² - 6x + 1) = 0

<=> 28x² - 30x + 2x⁴ - 12x³ = 0

<=> 2x(14x - 15 + x² - 6x²) = 0

<=> 2x(x² - 3x + 5)(x - 3) = 0

vì x² - 3x + 5 khác 0 nên:

<=> 2x = 0 hoặc x - 3 = 0

<=> x = 0 hoặc x = 3

\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)

\(\Leftrightarrow\frac{2x^4-12x^3+28x^2-30x}{2x^4-12x^3+28x^2-15x+2}=0\)

\(\Leftrightarrow2x^4-12x^3+28x^2-30x=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x^2-3x+5\right)=0\)

mà  \(x^2-3x+5\) khác 0 

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

có thể tách từng mẫu ra đi

Tách mẫu \(2x^2-6x+1\) ko đc

À,CHỈ CÓ 1 SỐ "0" THÔI NHÉ!

\(=>\frac{8}{2x^2-6x+2}-\frac{3}{2x^2-6x+2}=-1\)

\(=>\frac{5}{2x^2-6x+2}=-1\)

\(=>2x^2-6x+2=-5\)

\(=>2x^2-6x=-7\)

\(=>x.\left(2x-6\right)=-7\)

\(=>2x-6=-\frac{7}{x}\)

\(=>2x=\frac{-7+6x}{x}\)

\(=>3x=-7+6x\)

\(=>-7=-3x\)

\(=>x=\frac{-7}{-3}=\frac{7}{3}\)

E ms lớp 7 nên giải hơi dài thông cảm ạ :>

a) \(2x^3 + 6x^2 = x^2 +3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

S = \(\left\{-3;0;\dfrac{1}{2}\right\}\)

b) \((3x-1) (x^2 +2 ) = (3x-1) (7x - 10)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

S = \(\left\{\dfrac{1}{3};3;4\right\}\)

a)

voi x=0 ta thay 0 o phai la no pt

voi x<>0 chia ca 2 ve cho x^2 ta dc

x^2-3x+6-3/x+1/x^2=0

(x^2+1/x^2)-3(x+1/x)+6=0 dat a=x+1/x ta co (x+1/x)^2=a^2=>x^2+1/x^2=a^2-2

=>a^2-3a+4=0=>pt vo no :(