\(A=5-\sqrt{x+\sqrt{x}+1}\)
ĐK: \(x\ge0\)
=> \(x+\sqrt{x}\ge0\)
=> \(x+\sqrt{x}+1\ge1\)
=> \(\sqrt{x+\sqrt{x}+1}\ge1\)
=> \(-\sqrt{x+\sqrt{x}+1}\le1\)
Do đó: \(A\le4\)
Dấu "=" xảy ra khi x=0
\(B=\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+3}{1-\sqrt{x}}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{3x+6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi x=0
a)A= \(5-\sqrt{x+\sqrt{x}+1}\). ĐKXĐ: \(x\ge0\)
Ta luôn có: \(x+\sqrt{x}\ge0\) với \(x\ge0\)
\(\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x+\sqrt{x}+1}\ge1\)
\(\Rightarrow-\sqrt{x+\sqrt{x}+1}\le-1\)
\(\Rightarrow5-\sqrt{x+\sqrt{x}+1}\le4\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy GTLN của A=4 khi x=0
b) B= \(\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{1-\sqrt{x}}\). ĐKXĐ: \(x\ge0; x\ne1\)
= \(\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) = \(\frac{x+2\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) = \(\frac{\sqrt{x}+3}{\sqrt{x}+2}=\frac{\left(\sqrt{x+2}\right)+1}{\sqrt{x+2}}\)
= \(\frac{\sqrt{x}+2}{\sqrt{x}+2}+\frac{1}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)
Ta luôn có: \(\sqrt{x}+2\ge2\) với \(x\ge0; x\ne1\)
\(\Rightarrow\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\)
\(\Rightarrow1+\frac{1}{\sqrt{x}+2}\le\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy GTLN của B=\(\frac{3}{2}\) khi x=0