Question

rút gọn rồi tính giá trị biểu thức 

\(\dfrac{3x^2-12x+12}{x^2-4}\) tại x=\(-\dfrac{1}{4}\)

\(\dfrac{x^2-5x-6}{x^2-9}\) tại x=-1

 

\(\dfrac{x^2-9y^2}{x^2-6xy+9y^2}\) tại x=1, y=-\(\dfrac{2}{3}\)

Answer 1

a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)

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

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

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

\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)

b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)

\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)

=0