Question

thực hiện phép tính :

a) A= \(\dfrac{x^2-2x}{x^2+x-6}\times\dfrac{x^2-9}{x^2-3x}\)

b) B=\(\dfrac{2x^2+5x-3}{x^2-9}\times\dfrac{4x^2-1}{x^2-6x+9}\)

Answer 1

\(a,\)

\(A=\dfrac{x^2-2x}{x^2+x-6}.\dfrac{x^2-9}{x^2-3x}\)

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

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

\(b,\)

\(B=\dfrac{2x^2+5x-3}{x^2-9}.\dfrac{4x^2-1}{x^2-6x+9}\)

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

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

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