Question

Cho A= \(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

a) tìm đkxđ

b) tìm x để A=\(\dfrac{1}{5}\)

c) tìm gtnn của A

d) tìm x∈ Z để A nguyên

e) tìm X hữu tỉ để A∈Z

Answer 1

a, ĐKXĐ: x\(\ge0;x\ne1;x\ne9\)

b, A=\(\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{1}{5}\)

\(\Rightarrow5\sqrt{x}-15-\sqrt{x}-1=0\)

\(\Leftrightarrow4\sqrt{x}-16=0\)

\(\Leftrightarrow\sqrt{x}-4=0\)

\(\Leftrightarrow x=16\)

Answer 2

d: Để A là số nguyên thì \(\sqrt{x}+1-4⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{0;1;9\right\}\)

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

\(0< \dfrac{4}{\sqrt{x}+1}< =4\)

\(\Leftrightarrow0>-\dfrac{4}{\sqrt{x}+1}>=-4\)

=>1>A>=-3

A=0 khi căn x-3=0

=>x=9

A=-1 khi \(\sqrt{x}-3=-\sqrt{x}-1\)

=>\(2\sqrt{x}=2\)

=>x=1

A=-2 khi \(-2\sqrt{x}-2=\sqrt{x}-3\)

=>-3 căn x=-1

=>x=1/9

A=-3 khi \(-3\sqrt{x}-3=\sqrt{x}-3\)

=>x=0