`a) x^3 - 9x^2 + 14x = 0`

\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`

\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`

\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`

\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`

\(\Rightarrow\) `x.(x-7)(x-2)=0`

\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)

Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)

`3.x^3 - 5x^2 + 8x - 4 = 0`

\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`

\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`

\(\Rightarrow\) `(x^2 - 4x + 4)(x-1)=0`

\(\Rightarrow\) `(x-2)^2(x-1)=0`

\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)

Vậy \(x\in\left\lbrace2;1\right\rbrace\)

\(\Leftrightarrow-2x+1-x-2=8\cdot\left(-4x^2+6x-2x\right)+4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow-3x-1+32x^2-48x+16x-4x^2+8x-4=0\)

\(\Leftrightarrow28x^2-27x-5=0\)

\(\text{Δ}=\left(-27\right)^2-4\cdot28\cdot\left(-5\right)=1289>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{27-\sqrt{1289}}{56}\\x_2=\dfrac{27+\sqrt{1289}}{56}\end{matrix}\right.\)

a)

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

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

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

Vậy x = 2 ; x = - 1

b)

\(x^3+x^2+x+1=0\)

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

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

Vì x²+1 > 0

=> x + 1 = 0

=> x = - 1

Vậy x = - 1

c)

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

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

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

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

Vậy x = 1 ; x = - 3

d)

\(2x\left(3x-5\right)=10-6x\)

\(\Leftrightarrow2x\left(3x-5\right)+2\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{1}{2}\end{array}\right.\)

Vậy x = 5 / 3 ; x = - 1 / 2

để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì

x³+x²-x-1=0

=>(x³+x²)-(x+1)=0

=>x²(x+1)-(x+1)=0

=>(x+1)(x²-1)=0

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

=>(x+1)²(x-1)=0

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

vậy x=-1 hoặc x=1

1/ x² - 5x + 6 = 0 
⇔ x² - 2x - 3x + 6 = 0 
⇔ x(x - 2) - 3(x - 2) = 0 
⇔ (x - 2)(x - 3) = 0 
⇒S = {2 ; 3}.

1) \(x^2+5x+6=0\)

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

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

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

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

2) \(2\left(x+3\right)-x^2-3x=0\)

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

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

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

3) \(x^2+4x+3=0\)

\(\Leftrightarrow x^2+x+3x+3=0\)

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

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

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

4) \(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)

cái bài 2 câu 1 câu 2 và câu 3 sửa cái vế phải lại thành 3/2-1-2x/4 và -15/5 và 2.(x-1)/5