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Lời giải:
\(\sqrt{20152015}+\sqrt{20152017}-2\sqrt{20152016}=(\sqrt{20152015}-\sqrt{20152016})+(\sqrt{20152017}-\sqrt{20152016})\)
\(=\frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}\)
Dễ thấy: $0< \sqrt{20152015}+\sqrt{20152016}<\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{1}{\sqrt{20152015}+\sqrt{20152016}}>\frac{1}{\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}< 0$
$\Rightarrow \sqrt{20152015}+\sqrt{20152017}< 2\sqrt{20152016}$
Lời giải:
Ta có:
$\sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})$
Mà:
$(\sqrt{2015}+\sqrt{2017})^2=4032+2\sqrt{2015.2017}$
$=4032+2\sqrt{(2016-1)(2016+1)}=4032+2\sqrt{2016^2-1}$
$< 4032+2\sqrt{2016^2}=4.2016$
$\Rightarrow \sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}$
$\Rightarrow \sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})< \sqrt{2015}.2\sqrt{2016}$
Vậy......
đkxđ \(x\ne1;x\ge0\)
\(P=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2}{\left(\sqrt{x}\right)^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{1}{\sqrt{x}-1}-\frac{x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}+1-x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}+2}{\left(\sqrt{x}\right)^3-1}\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a.
\(A=\left[\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right].\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+x-\sqrt{x}-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2(x-2\sqrt{x}+1)}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2}{x+\sqrt{x}+1}\)
b.
Ta thấy với $x\geq 0 ; x\neq 1$ thì $x+\sqrt{x}+1\geq 1$
$\Rightarrow A=\frac{2}{x+\sqrt{x}+1}\leq 2$
Vậy $A$ đạt max bằng $2$ khi $x=0$
\(P=\left[\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{y-x}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}.\frac{\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\frac{x+2\sqrt{xy}+y-\sqrt{xy}}{x-2\sqrt{xy}+y+\sqrt{xy}}\)
\(=\frac{x+\sqrt{xy}+y}{x-\sqrt{xy}+y}\)
\(\left(\frac{x\sqrt{x}+x+2}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{x\sqrt{x}-x}\)
\(=\left(\frac{x\sqrt{x}+x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\frac{x\sqrt{x}-x}{1}\)
\(=\frac{x\sqrt{x}+x+2-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}-1\right)}{1}\)
\(=\frac{x\sqrt{x}+x-\sqrt{x}+3}{\sqrt{x}+1}.x\)
\(=\frac{x^2\sqrt{x}+x^2-x\sqrt{x}+3x}{\sqrt{x}+1}\)
\(........?!\)
