Question

Tìm x,y để biểu thức có giá trị bằng 0:

a) \(\dfrac{x^{2^{ }}-2x+1}{x^{2^{ }}-1}\)

b) \(\dfrac{x^{2^{ }^{ }}+x-2}{x-1}\)

Answer 1

a/ Ta có :

\(\dfrac{x^2-2x+1}{x^2-1}\) (Đktc \(x\ne\pm1\))

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

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

Mà : \(\dfrac{x^2-2x+1}{x^2-1}=0\)

\(\Leftrightarrow\dfrac{x-1}{x+1}=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy...

b/ \(\dfrac{x^2+x-2}{x-1}\) (đktc \(x\ne1\))

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

\(=x+2\)

Mà : \(\dfrac{x^2+x-2}{x-1}=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy..

Answer 2

câu b) là \(\dfrac{x^{2^{ }}+x-2}{x-1}\)