a)\(4x^2-7x-2=0\Leftrightarrow4x^2+x-8x-2=0\Leftrightarrow x\left(4x+1\right)-2\left(4x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{1}{4}\end{array}\right.\)

b)\(3x^2+10x+3=0\Leftrightarrow3x^2+9x+x+3=0\Leftrightarrow3x\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}3x+1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{3}\\x=-3\end{array}\right.\)

c)\(x^2-x-20=0\Leftrightarrow x^2+4x-5x-20=0\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)

d)\(6x^2+7x-3=0\Leftrightarrow6x^2-2x+9x-3=0\Leftrightarrow2x\left(3x-1\right)+3\left(3x-1\right)=0\)

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

e)\(10x^2-14x-12=0\Leftrightarrow2\left(5x^2-7x-6\right)=0\Leftrightarrow5x^2-7x-6=0\)

\(\Leftrightarrow5x^2+3x-10x-6=0\Leftrightarrow x\left(5x+3\right)-2\left(5x+3\right)=0\Leftrightarrow\left(x-2\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{3}{5}\end{array}\right.\)

7x² - 15x + 8 = 0

\(\Leftrightarrow\)7x² - 7x - 8x +8 = 0

\(\Leftrightarrow\)7x.(x - 1) - 8.(x - 1) = 0

\(\Leftrightarrow\)(7x - 8)(x - 1) = 0

\(\Leftrightarrow\)7x - 8 = 0 và x - 1 = 0

\(\Leftrightarrow\) x = 8/7 và x= 1

       x² - 5x - 6 = 0

<=>x² - x + 6x - 6 = 0

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

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

<=> x+6 = 0 và x-1 = 0

<=> x = -6, x= 1

a ) Ta có :   \(x^2-10+16=0\)

\(\Rightarrow x^2-10=-16\)

\(\Rightarrow x^2=-6\)

Mà \(x^2\ge0\forall x\Rightarrow x^2-10+16\)không có nghiệm 

b )  \(x^3+7x^2+2x-10=0\)

\(\Rightarrow x^3+7x^2+2x=10\)

\(\Rightarrow x.\left(x^2+7x+2\right)=10\)

\(\Rightarrow x=10\)

Làm tiếp nhé !!! 

c )   \(-3x^3+5x^2-8=0\)

\(\Rightarrow-3x^3+5x^2=8\)

\(\Rightarrow x^2.\left(-3x+5\right)=8\)

\(\Rightarrow x=...\)

\(=5x^2-10x+5+10x^2-10x=5\left(x^2-2x+1\right)+10x\left(x-1\right)\)

\(=5\left(x-1\right)^2+10x\left(x-1\right)=\left(x-1\right)\left(5\left(x-1\right)+10x\right)\)

\(=\left(x-1\right)\left(5x-5+10x\right)=\left(x-1\right)\left(15x-5\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

    \(15x-5=0\Rightarrow15x=5\Rightarrow x=\frac{5}{15}=\frac{1}{3}\)

vậy 1 và 1/3 là nghiệm của 15x^2-20x+5

\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)}=\dfrac{x+1}{2}\)

\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{7x\left(x+1\right)^2}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)}{2}\)

a, 15x - 6 = 12x + 3

\(\Leftrightarrow\) 15x - 12x = 3 + 6

\(\Leftrightarrow\) 3x = 9

\(\Leftrightarrow\) x = 3

Vậy S = {3}

b, \(\frac{x+2}{2}-\frac{2x-3}{5}=10x+\frac{13}{10}\)

\(\Leftrightarrow\) \(\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}=\frac{100x}{10}+\frac{13}{10}\)

\(\Leftrightarrow\) 5(x + 2) - 2(2x - 3) - 100x - 13 = 0

\(\Leftrightarrow\) 5x + 10 - 4x + 6 - 100x - 13 = 0

\(\Leftrightarrow\) -99x + 3 = 0

\(\Leftrightarrow\) x = \(\frac{1}{33}\)

Vậy S = {\(\frac{1}{33}\)}

d, (3x + 2)(4x - 5) = 0

\(\Leftrightarrow\) 3x + 2 = 0 hoặc 4x - 5 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) và x = \(\frac{5}{4}\)

Vậy S = {\(\frac{-2}{3}\); \(\frac{5}{4}\)}

Phần c với phần e bạn viết vậy mình ko hiểu, bn viết lại đi!

Chúc bn học tốt!!

a) 3x – 6 + x(x – 2) = 0

=> 3x - 6 + x² - 2x = 0

=> ( 3x - 2x ) - 6 + x² = 0

=> x - 6 + x² = 0

=> x² + x = 6

=> x( x + 1 ) = 2 . 3

=> x = 2

b) 2x(x – 3) – x(x – 6) – 3x = 0

=> 2x² - 6x - x² + 6x - 3x = 0

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

=> x² - 3x = 0

=> x( x - 3 ) = 0

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

b) 2x(x - 3) - x(x - 6) - 3x = 0

<=> 2x² - 6x - x² + 6x - 3x = 0

<=> x² - 3x = 0 

<=> x(x - 3) = 0

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

Vậy \(x\in\left\{0;3\right\}\)là nghiệm đa thức

c) 2x(3x  - 5) - x(6x + 1) + 22 = 0

<=> 6x² - 10x - 6x² - x + 22 = 0

<=> -11x + 22 = 0

<=> 11x = 22

<=> x = 2