Bài 8:
a: \(P=\left(1-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=\sqrt{2022+4\sqrt{2018}}-\sqrt{2022-4\sqrt{2018}}\)
\(=\sqrt{\left(\sqrt{2018}+2\right)^2}-\sqrt{\left(\sqrt{2018}-2\right)^2}\)
\(=\sqrt{2018}+2-\sqrt{2018}+2=4\)
Thay x=4 vào P, ta được:
\(P=\dfrac{2+1}{2}=\dfrac{3}{2}\)
Bài 10:
a: \(A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\)
\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{x}{\sqrt{x}+2}\right)\)
\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
b: \(A>=\dfrac{1}{3\sqrt{x}}\)
=>\(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}>=\dfrac{1}{3\sqrt{x}}\)
=>\(\dfrac{1}{\sqrt{x}+2}>=\dfrac{1}{3}\)
=>\(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{3}>=0\)
=>\(\dfrac{3-\sqrt{x}-2}{3\left(\sqrt{x}+2\right)}>=0\)
=>\(-\sqrt{x}+1>=0\)
=>\(-\sqrt{x}>=-1\)
=>\(\sqrt{x}< =1\)
=>0<=x<=1
Kết hợp ĐKXĐ, ta được: 0<x<=1
