Bài 9: Phân tích đa thức thành nhân tử

\(x^2-2 x y=x . x-2 x y=x(x-2 y)\)

a) \(6 y^3+2 y=2 y .\left(3 y^2+1\right)\)

b) \(4(x-y)-3 x(x-y)=(x-y)(4-3 x)\)

\(2 x^2+x=0 \Leftrightarrow x(2 x+1)=0 \Leftrightarrow\left[\begin{array}{c}x=0 \\ 2 x+1=0\end{array} \Leftrightarrow\left[\begin{array}{c}x=0 \\ x=\frac{-1}{2}\end{array}\right.\right.\)

Vậy \(x=0 ; x=\frac{-1}{2}\)

a) \((x+1)^2-y^2=(x+1+y)(x+1-y)\)
b) \(x^3+3 x^2+3 x+1=(x+1)^3\)
c) \(8 x^3-12 x^2+6 x-1=(2 x)^3-3 .2 x)^2 .1+3 . 2 x . 1-1^3=(2 x-1)^3\)

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

\(A=x^2+2 y-2 x-x y=\left(x^2+2 y\right)-(2 x+x y)=x(x+2)-x(2+y)=x[x+2-(2+y)]=x(x-y)\)Thay \(x=2022, y=2020\) vào A ta được:

\(A=2022 .(2022-2020)=2022.2=4044\)

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

Bạn Tròn có kết quả đúng, bạn vuông chưa phân tích triệt để.

a) \(x^2+x y=x . x+x . y=x(x+y)\)
b) \(6 a^2 b-18 a b=6 a b(a-3)\);
c) \(x^3-4 x=x\left(x^2-4\right)=x(x-2)(x+2)\);
d) \(x^4-8 x=x\left(x^3-8\right)=x(x-2)\left(x^2+2 x+4\right)\).

a) \(x^2-9+x y+3 y\)

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

\(=(x-3)(x+3)+y(x+3)\)

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

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

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

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

a) 

\(\begin{aligned} x^2-4 x=0 \\ \Leftrightarrow x(x-4)=0 \\ \Leftrightarrow\left[\begin{array}{c} x=0 \\ x-4=0 \end{array}\right. \\ \Leftrightarrow\left[\begin{array}{l} x=0 \\ x=4 \end{array}\right. \end{aligned}\)

Vậy \(x \in\{0 ; 4\}\)

b) 

\(\begin{aligned}2 x^3-2 x=0 \\ \Leftrightarrow 2 x\left(x^2-1\right)=0 \\\Leftrightarrow 2 x(x-1)(x+1)=0 \\\Leftrightarrow\left[\begin{array}{l} x=0 \\ x-1=0 \\ x+1=0 \end{array}\right. \\\Leftrightarrow\left[\begin{array}{l} x=0 \\ x=1 \\ x=-1 \end{array}\right. \\\end{aligned}\)

Vậy \(x \in\{0 ; 1 ;-1\}\)

a) \(S=x^2-(x-2 y)^2\) 

b) 

\( S=x^2-(x-2 y)^2=(x-x+2 y)(x+x-2 y)=2 y .(2 x-2 y) \)

\(=2 y .2(x-y)=4 y(x-y)\)

Khi \(x=102 \mathrm{~m}, y=2 m\) thì \(S=4.2 .(102-2)=800\left(\mathrm{~m}^2\right)\)