\(a,4x^2-25-\left(2x-5\right)\left(2x+7\right)\)

\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)

\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)

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

\(b,x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

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

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

=.= hok tốt!!

a) \(x^4+8x+63\)

\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)

\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)

c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)

Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)

\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)

\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)

\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)

a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)

= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+7x+10x\right)\)

= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

a) 3(x-y) - (x-y)^2

 =(x-y)(3-x+y)

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

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

\(a,=3\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(3-x+y\right)\\ b,=\left(x+y\right)^2-4x^2y^2=\left(x-2xy+y\right)\left(x+2xy+y\right)\\ c,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)

a, \(x^2\) + 4\(x\) - y² + 4

= (\(x^2\) + 4\(x\) + 4) - y²

= (\(x\) + 2)² - y²

= (\(x\) + 2 - y)(\(x\) + 2 + y)

b, 2\(x^2\) - 18

= 2.(\(x^2\) -9)

= 2.(\(x\) -3).(\(x\) + 3)

c, \(x^3\) - \(x^2\) - \(x\) + 1

= (\(x^3\) + 1) - (\(x^2\) +  \(x\))

= (\(x\) + 1)(\(x\)² - \(x\) + 1) - \(x\).(\(x\) + 1)

=(\(x\) + 1).(\(x^2\) - \(x\) + 1  - \(x\))

= (\(x\) + 1).(\(x\) - 1)²

1: =(x-1-y)(x-1+y)

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