Question

A= \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

a, rút gọn A

b, Tìm x để \(A\le0\)

Answer 1

hmm rút gọn nè :)))

\(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

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

Answer 2

b) Để \(A\le0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}\le0\)

rồi giải tiếp

Answer 3

a, A=\(\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{1}{\sqrt{x}-2}\)+\(\dfrac{1}{\sqrt{x}+2}\)

A=\(\text{​​}\text{​​}\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{1\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)+\(\dfrac{1\left(\sqrt{x-2}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

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

A=\(\dfrac{\sqrt{x}}{\sqrt{x-2}}\)

b,

Answer 4

ĐKXĐ:x\(\ge0\),x\(\ne4\)

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

b) Ta có \(A\le0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}\le0\)

Ta có \(\sqrt{x}\ge0\)

Suy ra \(\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

Kết hợp với ĐKXĐ

Vậy \(0\le x< 4\) thì A\(\le0\)