Question

Answer 1

a: \(P=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

b: \(P<\frac13\)

=>\(P-\frac13<0\)

=>\(\frac{\sqrt{x}-5}{\sqrt{x}+5}-\frac13<0\)

=>\(\frac{3\left(\sqrt{x}-5\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}<0\)

=>\(3\left(\sqrt{x}-5\right)-\sqrt{x}-5<0\)

=>\(2\sqrt{x}-20<0\)

=>\(\sqrt{x}<10\)

=>0<=x<100

Kết hợp ĐKXĐ, ta được: 0<=x<100 và x<>25

c: \(P-2=\frac{\sqrt{x}-5}{\sqrt{x}+5}-2=\frac{\sqrt{x}-5-2\sqrt{x}-10}{\sqrt{x}+5}=\frac{-\sqrt{x}-15}{\sqrt{x}+5}<0\)

=>P<2∀x thỏa mãn ĐKXĐ

d: Để P là số nguyên thì \(\sqrt{x}-5\vdots\sqrt{x}+5\)

=>\(\sqrt{x}+5-10\vdots\sqrt{x}+5\)

=>\(-10\vdots\sqrt{x}+5\)

=>\(\sqrt{x}+5\in\left\lbrace5;10\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;5\right\rbrace\)

=>x∈{0;25}

Kết hợp ĐKXĐ, ta được: x=0