Question

[Trần Ngọc Thảo]

Cho biểu thức

a) Rút gọn H

b) Tìm a để D < 2

c) Tính H khi a² + 3a = 0

d) Tìm a để H = 5

H = \(\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{\sqrt{2-\sqrt{a}}}\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge0;a\ne4\)

\(H=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

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

\(H< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}-2< 0\)

\(\Rightarrow\frac{\sqrt{a}-4-2\left(\sqrt{a}-2\right)}{\sqrt{a}-2}< 0\Rightarrow\frac{-\sqrt{a}}{\sqrt{a}-2}< 0\)

\(\Rightarrow\frac{\sqrt{a}}{\sqrt{a}-2}>0\Rightarrow\sqrt{a}-2>0\Rightarrow a>4\)

\(a^2+3a=0\Rightarrow a\left(a+3\right)=0\Rightarrow a=0\) (do \(a\ge0\Rightarrow a+3>0\))

\(\Rightarrow H=\frac{0-4}{0-2}=2\)

\(H=5\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\Rightarrow\sqrt{a}-4=5\sqrt{a}-10\)

\(\Rightarrow4\sqrt{a}=6\Rightarrow\sqrt{a}=\frac{3}{2}\Rightarrow a=\frac{9}{4}\)