\(2x^4-7x^3+17x^2-20x+14\)

\(=2x^4-3x^3+7x^2-4x^3+6x^2-14x+4x^2-6x+14\)

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

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

a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)

b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)

d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)

e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)

THAM KHẢO NHÉ BẠN, MÌNH KO VT ĐC TRONG NÀY NÊN PHẢI RA WORD VÍT R COPY VÀO Á:

$7x^2-13x+6=0\iff 7x^2-7x-6x+6=0$

$\iff 7x\left(x-1\right)-6\left(x-1\right)=0\iff \left(7x-6\right)\left(x-1\right)=0$

x=1;x=67

`#040911`

`7x^2 - 13x + 6`

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

`= (7x^2 - 7x) - (6x - 6)`

`= 7x(x - 1) - 6(x - 1)`

`= (7x - 6)(x - 1)`

\(7x^2-13x+6=7\left(x-1\right)\left(x-\dfrac{6}{7}\right)\)

=6x^2 - 4x - 9x +6

=(6x^2 -4x) - (9x-6)

=2x(3x -2) - 3(3x-2)

=(3x-2) (2x - 3)

1.

a) \(2x^4-4x^3+2x^2\)

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

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

b) \(2x^2-2xy+5x-5y\)

\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)

\(=2x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\cdot\left(2x+5\right)\)

2 . 

a,

\(4x\left(x-3\right)-x+3=0\)

⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)

⇒\(\left(x-3\right)\left(4x-1\right)=0\)

⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)

b, 

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

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

⇒\(\left(x-4\right)\left(3x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)

ban tích cho mk vs nha

bằng phương thức nào bạn ?

câu trả lời:

3x.x-13.x+6²

bằng phương pháp đặt nhân tử chung

\(=\left(13x^2-13x\right)-\left(14x-14\right)=13x\left(x-1\right)-14\left(x-1\right)=\left(13x-14\right)\left(x-1\right)\)

\(=6x^2+3x+10x+5=3x\left(2x+1\right)+5\left(2x+1\right)=\left(3x+5\right)\left(2x+1\right)\)

6x2+3x+10x+5=3x(2x+1)+5(2x+1)=(3x+5)(2x+1)