Question

Câu 1: dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng:

a) 5y/7=20xy/28x

b)3x(x+5)/2(x+5)=3x/2

c)x+2/x-1=(x+2)(x+1)/x²-1

d)x²-x-2/x+1=x²-3x+2/x-1

e)x³+8/x²-2x+4=x+2

Answer 1

a: \(\dfrac{20xy}{28x}=\dfrac{20xy:4x}{28x:4x}=\dfrac{5y}{7}\)

b: \(\dfrac{3x}{2}=\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)

c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)

d: \(\dfrac{x^2-x-2}{x+1}=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2\)

\(\dfrac{x^2-3x+2}{x-1}=\dfrac{\left(x-2\right)\left(x-1\right)}{x-1}=x-2\)

Do đó: \(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x^2-2x+4\right)}=x+2\)