Question

cho 2 biểu thức: A=\(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\) và B=\(\dfrac{2a+\sqrt{a}}{\sqrt{a}}-1\)

Rút gon biểu thức A-B

Answer 1

Ta có A-B=\(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\left(\dfrac{2a+\sqrt{a}}{\sqrt{a}}-1\right)=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}-\sqrt{a}}{\sqrt{a}}=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a}{\sqrt{a}}=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}=\dfrac{a^2+\sqrt{a}-2\sqrt{a}\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}=\dfrac{a^2+\sqrt{a}-2a\sqrt{a}+2a-2\sqrt{a}}{a-\sqrt{a}+1}=\dfrac{a^2-2a\sqrt{a}+2a-\sqrt{a}}{a-\sqrt{a}+1}=\dfrac{a^2-a\sqrt{a}+a-a\sqrt{a}+a-\sqrt{a}}{a-\sqrt{a}+1}=\dfrac{a\left(a-\sqrt{a}+1\right)-\sqrt{a}\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}=\dfrac{\left(a-\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a-\sqrt{a}+1}=a-\sqrt{a}\)