\(8a^4-2a^2-4a+2\)

\(=2\cdot\left(4a^4-a^2-2a+1\right)\)

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

\(8a^4-2a^2-4a+2\)

\(=2\left(4a^4-a^2-2a+1\right)\)

\(=2\left(4a^4-2a^3+2a^3-a^2-2a+1\right)\)

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

thêm bớt 4(x-a)^2 . a^2 là được

làm ra rứa ná

nói rõ ra đc ko bạn

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: \(2x^2-7xy+3y^2+x-3y\)

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

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

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

\(2a^2+8b^2-8ab\)

\(=2\left(a^2-4ab+4b^2\right)\)

\(=2\left(a-2b\right)^2\)

cám ơn nhaaaaa!!!!

2a2+8b2−8ab

=2(�2−4��+4�2)=2(a2−4ab+4b2)

=2(�−2�)2=2(a−2b)2

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(4a^3-3a+1\)

\(=\left(4a^3-4a\right)+\left(a+1\right)\)

\(=4a\left(a^2-1\right)+\left(a+1\right)\)

\(=4a\left(a-1\right)\left(a+1\right)+\left(a+1\right)\)

\(=\left(a+1\right)\left(4a^2-4a+1\right)\)

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

\(4a^2-4a+1-4b^2\)

<=>\(\left(2a-1\right)^2-4b^2\)

<=>\(\left(2a-1+2b\right)\left(2a-1-2b\right)\)

\(4a^2-4a+1-4b^2\)

\(=\left(2a-1\right)^2-4b^2\)

\(=\left(2a-1+2b\right)\left(2a-1-2b\right)\)

 4a^{2 -} 4b² - 4a -1 
= (4a²- 4a +1 ) - 4b²
= [(2a)² -2a.1 + 1² ] - (2b)2
= (2a -1 )² - (2b)2
= 2a - 1 - 2b ) . ( 2a - 1 + 2b )

f: Ta có: \(3a^3-15a+15b-3b^3\)

\(=3\left(a^3-5a+5b-b^3\right)\)

\(=3\left[\left(a-b\right)\left(a^2+ab+b^2\right)-5\left(a-b\right)\right]\)

\(=3\left(a-b\right)\left(a^2+ab+b^2-5\right)\)

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

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

x(x+1)(x+2)(x+3)+1

= [x(x+3)][(x+1)(x+2)]+1

=(x²+3x)(x²+3x+2)+1

Đặt x²+3x+1=y, ta có: 

(y-1)(y+1)+1

=y²-1+1

=y²

Thay y=x²+3x+1, lại có: 

(x²+3x+1)²