Question

MÌNH CẦN GẤP Ạ<

Answer 1

1: Thay x=16 vào A, ta được:

\(A=\frac{\sqrt{16}-1}{2\cdot\sqrt{16}}=\frac{4-1}{2\cdot4}=\frac38\)

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

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

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

3: \(P=A\cdot B\)

\(=\frac{\sqrt{x}-1}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}-1}{2\sqrt{x}-6}\)

Để P là số nguyên thì \(\sqrt{x}-1\) ⋮\(2\sqrt{x}-6\)

=>\(2\sqrt{x}-2\) ⋮\(2\sqrt{x}-6\)

=>\(2\sqrt{x}-6+4\) ⋮\(2\sqrt{x}-6\)

=>4⋮\(2\sqrt{x}-6\)

mà \(2\sqrt{x}-6\) ⋮2(x nguyên)

nên \(2\sqrt{x}-6\in\left\lbrace2;4;-2;-4\right\rbrace\)

=>\(2\sqrt{x}\in\left\lbrace8;10;4;2\right\rbrace\)

=>\(\sqrt{x}\) ∈{1;2;4;5}

=>x∈{1;4;16;25}