Question

Cho 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≠4)

a) Rút gọn P

b) tính giá tri của P với a=9

Answer 1

điều kiện xác định : \(x\ge0;x\ne4\)

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

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

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

\(\Leftrightarrow P=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

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

b) thế \(a=9\) vào \(P\) ta có : \(P=\dfrac{4}{\sqrt{9}-2}=\dfrac{4}{3-2}=4\)

Answer 2

Mysterious Person giúp e

Answer 3

DKXD: \(\left\{{}\begin{matrix}a\ge0\\a\ne4\end{matrix}\right.\)

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

= \(\dfrac{12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

b) \(\dfrac{12}{\left(\sqrt{9}-2\right)\left(\sqrt{9}+2\right)}=\dfrac{12}{5}\)