\(a^3+4a^2+4a+3\)

\(=a^3+3a^2+a^2+3a+a+3\)

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

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

thanks bạn

\(a^3+4a^2-7a-10\)

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

\(=\left(a^3+a^2\right)+\left(3a^2+3a\right)-\left(10a+10\right)\)

\(=a^2\left(a+1\right)+3a\left(a+1\right)-10\left(a+1\right)\)

\(=\left(a+1\right)\left(a^2+3a-10\right)\)

\(=\left(a+1\right)\left[\left(a^2+5a-2a-10\right)\right]\)

\(=\left(a+1\right)\left[a\left(a+5\right)-2\left(a+5\right)\right]\)

\(=\left(a+1\right)\left(a+5\right)\left(a-2\right)\)

a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

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

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

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

tự làm di

dễ đều có nhân tử chung là (x+1)

a) \(=\left(2x-1\right)^2\)

b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

a. \(4x^2-4x+1=\left(2x\right)^2-2x.2.1+1^2=\left(2x-1\right)^2\)

b. \(xy^2-x^3+2x^2-x=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x^2-2x+1\right)\right]=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

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

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

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

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

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

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

\(a,x-xy+y-y^2\\=(x-xy)+(y-y^2)\\=x(1-y)+y(1-y)\\=(1-y)(x+y)\\---\\b,x^2-4x-y+4(?)\\---\\c,x^2-2x-3\\=x^2+x-3x-3\\=x(x+1)-3(x+1)\\=(x+1)(x-3)\)

Bạn xem lại đề câu b nhé!

`x-xy+y-y^2`

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

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

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`x^2-4x-y+4`

`=(x^2-4x+4)-y`

`= (x-2)^2-y`

Thiếu đề?

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`x^2-2x-3`

`=x^2+x-3x-3`

`=(x^2+x)-(3x+3)`

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

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