= (x² - 2xy) - ( 5xy - 10y² )

=x(x-2y) - 5y( x-2y)

= (x-5y)(x-2y)

  x² - 7xy + 10y²

= x² - 2xy - 5xy + 10y²

= ( x² - 2xy ) - (5xy - 10y² )

= x ( x - 2y ) - 5y ( x-2y)

= ( x-2y ) ( x-5y )

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

=(2x-1)(2x+3)

bn làm đ rùi, tui đ cho bn

A=x¹⁴+x⁷+1

   =(x¹⁴+x¹³+x¹²)-(x¹³+x¹²+x¹¹)+(x¹¹+x¹⁰+x⁹)-(x¹⁰+x⁹+x⁸)+(x⁸+x⁷+x⁶)-(x⁶+x⁵+x⁴)+(x⁵+x⁴+x³)-(x³⁺x²+x)+(x²+x+1)

Đặt B=x²+x+1

=>A=x¹²B-x¹¹B+x⁹B-x⁸B+x⁶B-x⁴B+x³B-xB+B

=>A=B(x¹²-x¹¹+x⁹-x⁸+x⁶-x⁴+x³-x+1)

Thay B=x²+x+1 vào A là xong

dùng bơ du là ra chư mấy

Do lười đánh máy nên tớ mới đặt B khi làm khỏi đặt cũng dc cái này là toán lớp 8

Lời giải:
$2x^2+3xy-5y^2=(2x^2-2xy)+(5xy-5y^2)$

$=2x(x-y)+5y(x-y)=(x-y)(2x+5y)$

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

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

\(=x\left(x^2-8x+16\right)\)

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

c: Ta có: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

\(=2\left(x-y-3\right)\left(x-y+3\right)\)

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

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

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

f: Ta có: \(x^2-xy-2y^2\)

\(=x^2-2xy+xy-2y^2\)

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

\(=x^4-4x^2y^2+4y^4-9x^2y^2\)

\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)

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

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

\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)

\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

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

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

b: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)

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

Lời giải:
a.

Đặt $2a^2+5ab-3b^2-7b-2=(a+mb+n)(2a+pb+k)$ với $m,n,p,k$ nguyên 

$\Leftrightarrow 2a^2+5ab-3b^2-7b-2=2a^2+ab(2m+p)+mpb^2+a(k+2n)+b(km+np)+kn$ 

Đồng nhất hệ số:

\(\left\{\begin{matrix} 2m+p=5\\ mp=-3\\ k+2n=0\\ km+np=-7\\ kn=-2\end{matrix}\right.\)

Giải hpt này ta thu được $m=3; n=1; p=-1; k=-2$

Vậy $2a^2+5ab-3b^2-7b-2=(a+3b+1)(2a-b-2)$

b. Đa thức không phân tích được thành nhân tử

lm theo pp đồng nhất hệ số ạ

b: Ta có: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)

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