a: ĐKXĐ: x>0; x<>1
\(Q=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(=x-\sqrt{x}+1\)
b: \(Q=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu = xảy ra khi căn x-1/2=0
=>căn x=1/2
=>x=1/4
c: \(P=3\cdot\dfrac{Q}{\sqrt{x}}\)
\(=\dfrac{3x-3\sqrt{x}+3}{\sqrt{x}}\)
P là số nguyên
=>\(3x-3\sqrt{x}+3⋮\sqrt{x}\)
=>3 chia hết cho căn x
=>\(\sqrt{x}\in\left\{1;3\right\}\)
mà x<>1
nên căn x=3
=>x=9
