Question

[Vinh Thuy Duong]

Bài 3: Chứng minh các phân thức sau bằng nhau

a)\(\dfrac{x+1}{x+3}\)=\(\dfrac{x^2+4x+3}{x^2+6x+9}\)

b)\(\dfrac{x+y}{3x}\)=\(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)

Answer 1

[Nguyễn Huy Tú]

\(a,VP=\dfrac{x^2+4x+3}{x^2+6x+9}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x+1}{x+3}=VT\)

Vậy ta có đpcm 

b, \(VP=\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}=\dfrac{x+y}{3x}=VT\)

Vậy ta có đpcm 

Answer 2

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\dfrac{x^2+4x+3}{x^2+6x+9}\)

\(=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)}\)

\(=\dfrac{x+1}{x+3}\)

b: Ta có: \(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)

\(=\dfrac{3x\left(x+y\right)\left(x+y\right)}{3x\cdot3x\cdot\left(x+y\right)}\)

\(=\dfrac{x+y}{3x}\)