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![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge-1;y\ge0\)
\(x+y+\sqrt{8y}+5=4\sqrt{x+1}+\sqrt{2}\sqrt{xy+y}\)
\(\Leftrightarrow\)\(\left(x+1-4\sqrt{x+1}+4\right)-\left(\sqrt{x+1}\sqrt{2y}-2\sqrt{2y}\right)+y=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x+1}-2\right)^2-\sqrt{2y}\left(\sqrt{x+1}-2\right)+y=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x+1}-2\right)^2-2\sqrt{\frac{y}{2}}\left(\sqrt{x+1}-2\right)+\frac{y}{2}+\frac{y}{2}=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x+1}-\frac{y}{2}-2\right)^2+\frac{y}{2}=0\)
Có: \(\left(\sqrt{x+1}-\frac{y}{2}-2\right)^2+\frac{y}{2}\ge0\) ( do \(y\ge0\) )
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x+1}-\frac{y}{2}-2=0\\\frac{y}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
...
\(\frac{1}{x}+\frac{25}{y}\ge\frac{\left(1+5\right)^2}{x+y}\ge\frac{6^2}{6}=6\)
Dấu "=" xảy ra khi \(x+y=6\) và \(\frac{1}{x}=\frac{5}{y}=\frac{1+5}{x+y}=\frac{6}{6}=1\)\(\Rightarrow\)\(x=1;y=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)
\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)
\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)
\(=\left(4-8+25\right)\sqrt{x^2+1}\)
\(=21\sqrt{x^2+1}\)
b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)
\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)
\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)
\(B=\sqrt{3}\)