Question

Bài 1: Phân tích đa thức thành nhân tử
a) \(4x^2\)-16
b) \(4x\left(x-1\right)+y\left(1-x\right)\)

c) \(8x\left(x^2-2\right)-16x\)

d) \(\left(x-y\right)^2-4\left(x-y\right)\)
Bài 2: Tìm x

a) \(x^2\left(x-3\right)-4\left(x-3\right)=0\)

b) \(6x\left(x-1\right)-5x=0\)

Mong mn giúp ạ

Answer 1

Bài 2

a) \(x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\\x=3\end{matrix}\right.\)

b) \(6x\left(x-1\right)-5x=0\)

\(\Rightarrow6x^2-6x-5x=0\)

\(\Rightarrow6x^2-11x=0\)

\(\Rightarrow x\left(6x-11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\6x-11=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{6}\end{matrix}\right.\)

 

Answer 2

Bài 1

a) \(4x^2-16=\left(2x\right)^2-4^2=\left(2x+4\right)\left(2x-4\right)\)

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

c) \(8x\left(x^2-2\right)-16x=8x\left(x^2-2-2\right)=8x.x^2=8x^3\)

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