Question

[Phạm Phương Anh]

cần gấp nha mn 

Answer 1

[Yen Nhi]

Answer:

Bài 1:

\(x^2-xy+2x-2y\)

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

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

\(x^2-4y^2\)

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

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

Bài 2:

\(5\left(x-3\right)+x^2-3x=0\)

\(\Rightarrow5\left(x-3\right)+x\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5+x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)

\(x^2-x-12=0\)

\(\Rightarrow x^2+3x-4x-12=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)

Bài 3:

a, Với \(x=-3\)

\(A=\frac{x-4}{x+5}=\frac{-3-4}{-3+5}=-\frac{7}{2}\)

b, \(B=\frac{2}{x+4}+\frac{x+20}{x^2-16}\left(ĐK:x\ne\pm4;x\ne-5\right)\)

\(=\frac{2}{x+4}+\frac{x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{2\left(x-4\right)+x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{2x-8+x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3x+12}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3}{x-4}\)

c, \(M=A.B=\frac{x-4}{x+5}.\frac{3}{x-4}=\frac{3}{x+5}\)

Để M nguyên thì \(3⋮\left(x+5\right)\)

\(\Rightarrow x+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{-4;-6;-2;-8\right\}\) mà \(ĐK:x\ne\pm4\)

Vậy \(x\in\left\{-6;-2;-8\right\}\)