Question

[Bảo Thiii]

giúp e bài 13 14 vs ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 14:

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

b: \(A=\dfrac{x}{2x+4}+\dfrac{3x+2}{x^2-4}\)

\(=\dfrac{x}{2\left(x+2\right)}+\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x\left(x-2\right)+2\left(3x+2\right)}{2\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+4x+4}{2\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{2\left(x-2\right)}\)

c: Đặt B=2*A

\(\Leftrightarrow B=\dfrac{2\cdot\left(x+2\right)}{2\left(x-2\right)}=\dfrac{x+2}{x-2}\)

Để B là số nguyên thì \(x+2⋮x-2\)

=>\(x-2+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;1;4;0;6\right\}\)

Bài 13:

1:

a: \(\dfrac{x^2-y^2}{x^2+xy}\cdot\dfrac{x+2y}{x-y}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{x\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{x+2y}{x}\)

b: \(x^2\cdot\left(2x-3y^2\right)-4xy\left(1-xy\right)-2x^3\)

\(=2x^3-3x^2y^2-4xy+4x^2y^2-2x^3\)

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

2:

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

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

\(=3x^2-12x+8\)

\(f\left(4\right)=3\cdot4^2-4=48-4=44\)