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`

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=7\\x_1x_2=-6\end{matrix}\right.\)

\(E=2x_1^2x_2+2x_1x_2^2\\ =2x_1x_2\left(x_1+x_2\right)\\ =2.\left(-6\right).7\\ =-84\)

mình vừa lên lớp 9 , chưa học phương trình bậc 2 

a)2x³ + 7x² - x - 12 =0

=>2x³+x²-4x+6x²+3x-12=0

=>x(2x²+x-4)+3(2x²+x-4)=0

=>(x+3)(2x²+x-4)=0

=>x+3=0 hoặc 2x²+x-4=0

Xét x+3=0 <=>x=-3

Xét 2x²+x-4=0 ta dùng delta

\(\Delta=1^2-\left(-4\left(2.4\right)\right)=33>0\)

=>pt có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{33}}{4}\)

b)- x^3 + x^2 + 7x + 2 =0

=>-x³+3x²+x-2x²+6x+2=0

=>-x(x²-3x-1)+(-2)(x²-3x-1)=0

=>-(x+2)(x²-3x-1)=0

=>-(x+2)=0 hoặc x²-3x-1=0

Xét -(x+2)=0 <=>x=-2

Xét x²-3x-1=0 theo delta ta có:

\(\Delta=\left(-3\right)^2-\left(-4\left(1.1\right)\right)=13>0\)

=>pt cũng có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\)

xài hóc ne đi

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

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

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

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

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

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

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 .....