Lời giải:
$2x^2-7x+6=0$

$\Leftrightarrow (2x^2-4x)-(3x-6)=0$

$\Leftrightarrow 2x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(2x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $2x-3=0$

$\Leftrightarrow x=2$ hoặc $x=\frac{3}{2}$

2x² - 7x + 6 = 0

\(\Leftrightarrow\) 2x² - 4x - 3x + 6 = 0

\(\Leftrightarrow\) (2x² - 4x) - (3x - 6) = 0

\(\Leftrightarrow\) 2x(x - 2) - 3(x - 2) = 0

\(\Leftrightarrow\) (x - 2)(2x - 3) = 0

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

S = \(\left\{2,\dfrac{3}{2}\right\}\)

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

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

a)

`x^2 +5x+6=0`

`<=> x^2 + 3x +2x+6=0`

`<=> x(x+3)+2(x+3)=0`

`<=> (x+3)(x+2)=0`

`<=> x+3=0 hoặcx+2=0`

`<=> x=-3 hoặc x=-2`

b)

`x^2 -7x+6=0`

`<=> x^2 -6x-x+6=0`

`<=> x(x-6)-(x-6)=0`

`<=> (x-6)(x-1)=0`

`<=> x-6=0 hoặc x-1=0 `

`<=> x=6 hoặc x=1`

c)

`x^2 +x -12=0`

`<=> x^2 +4x-3x-12=0`

`<=> x(x+4)-3(x+4)=0`

`<=> (x+4)(x-3)=0`

`<=> x+4=0 hoặc x-3=0`

`<=> x=-4 hoặc x=3`

d)

`x^2 -x-6=0`

`<=>x^2 -3x+2x-6=0`

`<=> x(x-3)+2(x-3)=0`

`<=> (x-3)(x+2)=0`

`<=> x-3=0 hoặc x+2=0`

`<=> x=3 hoặc x=-2`

e)

`2x^2 -3x-5=0`

`<=> 2x^2 -5x+2x-5=0`

`<=> x(2x-5)+(2x-5)=0`

`<=> (2x-5)(x+1)=0`

`<=> 2x-5=0 hoặc x+1=0`

`<=> x=5/2 hoặc x=-1`

Chăm chỉ wa' ;-;

\(\Leftrightarrow x\left(5x^2-7x+5x-7\right)=0\\ \Leftrightarrow x\left[5x\left(x+1\right)+7\left(x+1\right)\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{7}{5}\end{matrix}\right.\)

\(\Leftrightarrow5x^3+5x^2-7x^2-7x=0\)

\(\Leftrightarrow5x^2\left(x+1\right)-7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x^2-7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\5x^2-7x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

a)Ta có \(\left(2x+1\right)\left(x^2+2\right)=0\)<=>

2x+1=0<=>x=\(-\frac{1}{2}\)

hoặc \(x^2+2=0\)<=>\(x^2=-2\)(Vô lí)

Vậy tập nghiệm của pt S=(\(-\frac{1}{2}\))

b)\(\left(x^2+4\right)\left(7x-3\right)=0\)

<=>\(\left[{}\begin{matrix}x^2+4=0\\7x-3=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x^2=-4\\x=\frac{3}{7}\end{matrix}\right.\)

\(x^2=-4\) vô lí

Vậy ..........

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

<=>\(\left[{}\begin{matrix}x^2+x+1=0\\6-2x=0\end{matrix}\right.\)

Vì \(x^2+x+1>0\)(dễ dàng c/m)

=>6-2x=0=>x=3

Vậy...

d)\(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

<=>8x-4=0,x=\(\frac{1}{2}\)

hoặc \(x^2+2x+2=0\)(vô lí)

Vậy .....

3)

\(x^3-7x+6=0\)

\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)

\(\Leftrightarrow\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(2x+6\right)=0\)

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

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

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

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

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

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

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

\(\Leftrightarrow3x\left(x+2\right)=0\)

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

Vậy ................

Lạnh xun loz

phải không mày >

haizz

\(6x^4+7x^3-36x^2-7x+6=0\)

\(\Leftrightarrow\left(6x^4-11x^3-3x^2+2x\right)+\left(18x^3-33x^2-9x+6\right)=0\)

\(\Leftrightarrow x\left(6x^3-11x^2-3x+2\right)+3\left(6x^3-11x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(6x^3-11x^2-3x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\left(6x^3-14x+4x\right)+\left(3x^2-7x+2\right)\right]\left(x+3\right)=0\)

\(\Leftrightarrow\left[2x\left(3x^2-7x+2\right)+\left(3x^2-7x+2\right)\right]\left(x+3\right)=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-1=0\end{cases}}\)hoặc \(\orbr{\begin{cases}2x+1=0\\x+3=0\end{cases}}\)

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

Vậy tập hợp nghiệm \(S=\left\{2;-3;\frac{1}{3};\frac{-1}{2}\right\}\)

a)\(9x^2+5x+2=0\)

\(\Delta=5^2-4\cdot9\cdot2=-47< 0\)

Vô nghiệm

b)\(5x^2+4x-2=0\)

\(\Delta=4^2-4\cdot5\cdot\left(-2\right)=56\)

\(x_{1,2}=\frac{-4\pm\sqrt{56}}{10}\)

c)\(2x^3+7x^2+7x+2=0\)

\(\Rightarrow2x^3+6x^2+4x+x^2+3x+2=0\)

\(\Rightarrow2x\left(x^2+3x+2\right)+\left(x^2+3x+2\right)=0\)

\(\Rightarrow\left(x^2+3x+2\right)\left(2x+1\right)=0\)

\(\Rightarrow\left(x^2+2x+x+2\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[x\left(x+2\right)+\left(x+2\right)\right]\left(2x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(2x+1\right)=0\)

=>x=-1 hoặc x=-2 hoặc \(x=-\frac{1}{2}\)