Question

Cho biểu thức :\(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)

(a>0;a khác 0)

a)Rút gọn P

b) Tính giá trị P khi a=\(6-2\sqrt{5}\)

Answer 1

a)P= \(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)

=\(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{a-4}\)

=\(\dfrac{a+3\sqrt{a}+2\sqrt{a}+6-a+\sqrt{a}+2\sqrt{a}-2-4\sqrt{a}+4}{a-4}\)

=\(\dfrac{4\sqrt{a}+8}{a-4}=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\dfrac{4}{\sqrt{a}-2}\)

b) thay a=\(6-2\sqrt{5}\)vào P ta có :

\(\dfrac{4}{\sqrt{6-2\sqrt{5}}-2}=\dfrac{4}{\sqrt{\left(\sqrt{5}-1\right)^2}-2}=\dfrac{4}{\sqrt{5}-3}\)