Question

Cho P = \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

a) Tính P khi x = 6 - \(2\sqrt{5}\)

b) Tìm GTNN của P khi x > 25

c) Tìm x để P < \(\sqrt{x}\)

d) Tìm x nguyên để Q = \(\frac{P}{\sqrt{x}}\) nhận GT nguyên

giúp mình với ạ, mình camon nhèo <3

Answer 1

ĐKXĐ: ...

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

\(=\frac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\left(5-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{-\sqrt{x}}{\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(5-\sqrt{x}\right)}=\frac{x}{\sqrt{x}-5}\)

\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\Rightarrow x=\sqrt{5}-1\)

\(\Rightarrow P=\frac{6-2\sqrt{5}}{\sqrt{5}-6}=...\)

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

\(\Rightarrow P\ge2\sqrt{\frac{25\left(\sqrt{x}-5\right)}{\sqrt{x}-5}}+10=20\)

\(\Rightarrow P_{min}=20\) khi \(x=100\)

\(P< \sqrt{x}\Rightarrow\frac{x}{\sqrt{x}-5}< \sqrt{x}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-5}< 1\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-5}-1< 0\)

\(\Rightarrow\frac{5}{\sqrt{x}-5}< 0\Rightarrow\sqrt{x}-5< 0\Rightarrow x< 25\)

Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0< x< 25\\x\ne3\\\end{matrix}\right.\)

\(Q=\frac{\sqrt{x}}{\sqrt{x}-5}=1+\frac{5}{\sqrt{x}-5}\)

Để Q nguyên \(\Rightarrow\sqrt{x}-5=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow\sqrt{x}=\left\{0;4;6;10\right\}\Rightarrow x=\left\{16;36;100\right\}\)