Question

Giúp mình nha.Chi tiết 1 chút nha..

Answer 1

a: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)

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

b: \(P=x-\sqrt{x}\)

\(=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>x=1/4(nhận)