\(=x^2-x+2022x-2022\\ =x\left(x-1\right)+2022\left(x-1\right)\\ =\left(x+2022\right)\left(x-1\right)\)

x⁴+2012x²+2012x+2012

=(x⁴-x)+(2012x²+2012x+2012)

=x(x³-1)+2012(x²+x+1)

=x(x-1) (x²+x+1) + 2012 (x²+x+1)

=(x²+x+1) [x(x-1)+2012]

=(x²+x+1) (x²-x+2012)

x⁴ + 2021x² - 2020x + 2021

= (x⁴ + x) + 2021(x² - x + 1)

= x(x³ + 1) + 2021(x² - x + 1)

= x(x + 1)(x² - x + 1) + 2021(x² - x + 1)

= (x² + x + 2021)(x² - x + 1)

Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

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

pls help me mk đang cần vội :(

Bài 1:

\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)

Bài 2:

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

Bài 2: 

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

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

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

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

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

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

c)

\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)

\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)

\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)

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

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

b)

\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)

\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)

\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)

CÓ:   \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)

=>   \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)

=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))

Bài 1:

a: \(5x^3+10xy=5x\left(x^2+2y\right)\)

b: \(x^2+14x+49-y^2\)

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

\(=\left(x+7+y\right)\left(x+7-y\right)\)

thử dùng hệ số bất định xem thanh niên

hệ số bất định là cqq j?

ở đây ko cho cop link, lên mạng xem thử ik~