Question

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

Tìm các số nguyên cua x để A nhận giá trị nguyên         

Answer 1

\(A=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\)

A nguyên khi \(\frac{3}{\sqrt{x}-2}\) nguyên

<=> \(\sqrt{x}+2\) là ước của 3

\(\Leftrightarrow\sqrt{x}+2\in\left\{1;3;-1;-3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-1;1;-3;-5\right\}\)

\(\Leftrightarrow x\in\left\{1;2;9;25\right\}\)

Vậy x=1 ; x= 9 ; x=25

Answer 2

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}-2}+\frac{3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow3⋮\sqrt{x}-2\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;5;-1\right\}\)

\(\Rightarrow x\in\left\{9;1;25\right\}\)

Answer 3

\(A=\frac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\frac{3}{\sqrt{x}-2}\)

A nguyên <=> \(\frac{3}{\sqrt{x}-2}\)

<=> \(\sqrt{x}-2\inƯ_3\)

<=>\(\sqrt{x}-2\in\left\{1;3;-1;-3\right\}\)

<=>\(\sqrt{x}\in\left\{3;5;1;-1\right\}\)

=> \(x\in\left\{5;25;1;1\right\}\)

Vậy \(x\in\left\{5;25;1\right\}\)

Answer 4

A =  \(\frac{\sqrt{X}+1}{\sqrt{X}-2}\)

=\(\frac{\sqrt{X}-2+3}{\sqrt{X}-2}\)

=\(1+\frac{3}{\sqrt{X}-2}\)

A nhận giá trị nguyên \(\Leftrightarrow\)\(\frac{3}{\sqrt{x}-2}\)nhận giá trị nguyên

\(\Leftrightarrow\)\(\sqrt{x}-2\inƯ\left(3\right)\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;3;-3\right\}\)

+) \(\sqrt{x}-2=1\Leftrightarrow x=9\)

+)\(\sqrt{x}-2=-1\Leftrightarrow x=1\)

+) \(\sqrt{x}-2=3\Leftrightarrow x=25\)

+)\(\sqrt{x}-2=-3\Leftrightarrow x=1\)

vậy   x=1; x=25; x=9 thì A nhận giá trị nguyên

Answer 5

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)+3}{\sqrt{x}-2}=\sqrt{x}-2+\frac{3}{\sqrt{x}-2}\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\)

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