Question

Answer 1

Bài 1:

1) 20x²y-8xy³=4xy(5x-2y²)

2) 16x³y²+20xyz=4xy(4x²y+5z)

3) 10(x-y²)-6y(x-y²)=(10-6y)(x-y²)

4) 3x(z+2)+7(-z-2)=3x(z+2)-7(z+2)=(3x-7)(z+2)

5) (5-x)³-(x-5)=(5-x)³+(5-x)=(5-x)\(\left[\left(5-x\right)^2+1\right]\)=(5-x)(25-10x+x²+1)=(5-x)(26-10x+x²)

Answer 2

Bài 2:

a) 9x²-15x³=0

⇒3x²(3-5x)=0

\(\Rightarrow\left[{}\begin{matrix}3x^2=0\\3-5x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{5}\end{matrix}\right.\)

b) 3x(x+6)-2(x+6)=0

\(\Rightarrow\left(3x-2\right)\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-6\end{matrix}\right.\)

c) 4x(x-2021)+2021-x=0

⇒4x(x-2021)+(2021-x)=0

⇒4x(x-2021)-(x-2021)=0

⇒(4x-1)(x-2021)=0

\(\Rightarrow\left[{}\begin{matrix}4x-1=0\\x-2021=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=2021\end{matrix}\right.\)

d) 15x(x-2)+4(2-x)=0

⇒15x(x-2)-4(x-2)=0

⇒(x-2)(15x-4)=0

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\15x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{15}\end{matrix}\right.\)

e) (x-1)-(x-1)²=0

⇒(x-1)(1-x+1)=0

⇒(x-1)(2-x)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)