\(ĐKXĐ:x\ge\frac{1}{2}\)

Phương trình đã cho tương đương :

\(4.\left(x^2+1\right)+3.x.\left(x-2\right).\sqrt{2x-1}=2x^3+10x\)

\(\Leftrightarrow3x\left(x-2\right)\sqrt{2x-1}=2x^3-8x^2+10x-4\)

\(\Leftrightarrow3x.\left(x-2\right).\sqrt{2x-1}=2.\left(x-2\right).\left(x-1\right)^2\) (1)

Dễ thấy \(x=2\) là một nghiệm của (1). Xét \(x\ne2\). Khi đó ta có :

\(3x.\sqrt{2x-1}=2.\left(x-1\right)^2\)(*)

Đặt \(\sqrt{2x-1}=a\left(a\ge0\right)\Rightarrow-a^2=1-2x\)

Khi đó pt (*) có dạng :

\(3x.a=2.\left(x^2-a^2\right)\)

\(\Leftrightarrow2x^2-3xa-2a^2=0\)

\(\Leftrightarrow2x^2-4ax+xa-2a^2=0\)

\(\Leftrightarrow2x.\left(x-2a\right)+a.\left(x-2a\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2a=x\\a=-2x\end{cases}}\)

+) Với \(2a=x\Rightarrow2\sqrt{2x-1}=x\left(x\ge0\right)\)

\(\Leftrightarrow x^2=4.\left(2x-1\right)\)

\(\Leftrightarrow x^2-8x+4=0\)

\(\Leftrightarrow x=4\pm2\sqrt{3}\) ( Thỏa mãn )

+) Với \(a=-2x\Rightarrow\sqrt{2x-1}=-2x\left(x\le0\right)\)

\(\Leftrightarrow4x^2=2x-1\)

\(\Leftrightarrow4x^2-2x+1=0\) ( Vô nghiệm )

Vậy phương trình đã cho có tập nghiệm \(S=\left\{4\pm2\sqrt{3},2\right\}\)

Đặt \(2x^2-2x+2=a\)

\(\Leftrightarrow\dfrac{a-3x}{x-1}=\dfrac{a+3x+15}{x-3}\)

\(\Leftrightarrow6x^2+3x+2a-15=0\)

\(\Leftrightarrow10x^2-x-11=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=-1\end{matrix}\right.\)

a/ \(2x-3=5x+2\)

\(\Leftrightarrow5x-2x=-3-2\)

\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)

Vậy..

b. \(2x\left(x-1\right)=2x+2\)

\(\Leftrightarrow2x^2-4x-2=0\)

\(\Leftrightarrow x^2-2x-1=0\)

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

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

Vậy...

c/ ĐKXĐ : \(x\ne\pm2\)

\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)

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

\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)

\(\Leftrightarrow2x-16=0\)

\(\Leftrightarrow x=8\)

Vậy..

Phần b bằng bn vậy ? 

a/ \(2x-3=5x+2\)

\(< =>2x-5x=3+2\)

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

\(< =>x=-\dfrac{5}{3}\)

Vậy.....

b/ ( đàu bài thiếu)

c/ \(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{12}{2\left(x+2\right)}\)

ĐKXĐ của phương tình là: \(x\ne\pm2\)

\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{12}{2\left(x+2\right)}\)

\(< =>\dfrac{2\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{12\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)

\(< =>\left(2x+4\right)\left(x+2\right)-2x^2-12x+24=0\)

\(< =>2x^2+4x+4x+8-2x^2-12x+24=0\)

\(< =>-4x+32=0\)

\(< =>-4x=-32\)

\(< =>x=-32:-4=7\)

Vậy....

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

a: Đặt x-3=a; x+1=b

Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>(x-3)(x+1)(2x-2)=0

hay \(x\in\left\{3;-1;1\right\}\)

b: \(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-15x^2-9x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)^2+6x\left(2x^2+1\right)-4x\left(2x^2+1\right)-24x^2=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(2x^2+6x+1\right)-4x\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2+3x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x^2+3x+\dfrac{9}{4}=\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

hay \(x\in\left\{\dfrac{\sqrt{7}-3}{2};\dfrac{-\sqrt{7}-3}{2}\right\}\)