a: =(x+6)(x-1)

n: \(=4x^4+36x^2+81-36x^2\)

\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)

giúp me

\(=\left(m-n\right)\left(m+n\right)-2\left(m+n\right)=\left(m+n\right)\left(m-n-2\right)\)

\(m^2-n^2-2m-2n=\left(m^2-n^2\right)-\left(2m+2n\right)=\left(m-n\right)\left(m+n\right)-2\left(m+n\right)=\left(m+n\right)\left(m-n-2\right)\)

Đặt n^2 + 2n + 1= a, ta được:

(a - 1)(a + 1) +1= a^2  -  1 + 1= a^2=(n^2 + 2n +1)^2

=(n + 1)^4

f, \(a^2-5a+14\) không phân tích được thành nhân tử.

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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

g) 2(m+1)(m+4)

g) `2m^2+10m+8`

`=(2m^2+2m)+(8m+8)`

`=2m(m+1)+8(m+1)`

`=(m+1)(2m+8)`

f) `4p^2-36p+56`

`=(4p^2-8p)-(28p-56)`

`=4p(p-2)-14(p-2)`

`=(p-2)(4p-14)`

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

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)