Về mở sách học lại 7 hằng đẳng thức đi nha bạn. Kiến thức cơ bản không làm được là mai sau thi cấp 3 trượt thẳng cổ đấy 

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

⇔(2x+1−x+1)(2x+1+x−1)

⇔(x+2)3x

1.

= (x^3 + 125 ) -(x^2 +5x)

=(x +5) (x^2 -5x +25) -x(x+5)

=(x+5)(x^2 -5x +25 -x)

=(x+5)(x^2 -6x +25)

2.

= (x^3 -27) + (2x^2 -6x)

=(x-3) (x^2 +3x +9) +2x (x-3)

=(x-3) (x^2 +3x +9 +2x)

=(x-3) (x^2 +5x +9)

x⁴ + x³ + 2x² + x + 1 

= (x⁴ + 2x² + 1) + (x³ + x)

= (x² + 1)² + x (x² + 1)

= (x² + 1) ( x² + 1 + x)

= (x² + 1) (x + 1)²

Trả lời:

x⁴ + x³ + 2x² + x + 1

= ( x⁴ + 2x² + 1 ) + ( x³ + x )

= ( x² + 1 )² + x ( x² + 1)

= ( x² + 1 ) ( x² + 1 + x )

= ( x² + 1) ( x + 1 )²

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

Là nhân tử rồi bn ơi

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

`(x-1)^2-2(x-1)(2x+1)+(2x+1)^2`

`=(x-1-2x-1)^2`

`=(-x-2)^2`

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

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