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1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)
a) \(\dfrac{x^2+2}{\sqrt{x^2+1}}\ge2\) \(\Leftrightarrow\) \(x^2+2\ge2\sqrt{x^2+1}\)
\(\Leftrightarrow\) \(\left(x^2+2\right)^2\ge\left(2\sqrt{x^2+1}\right)^2\) \(\Leftrightarrow\) \(x^4+4x^2+4\ge4x^2+4\)
\(\Leftrightarrow\) \(x^4\ge0\) (đúng \(\forall x\)) \(\Rightarrow\) \(\dfrac{x^2+2}{\sqrt{x^2+1}}\ge2\) (đpcm)
b) \(\dfrac{2x^2+1}{\sqrt{4x^2+1}}\ge1\) \(\Leftrightarrow\) \(2x^2+1\ge\sqrt{4x^2+1}\)
\(\Leftrightarrow\) \(\left(2x^2+1\right)^2\ge\left(\sqrt{4x^2+1}\right)^2\) \(\Leftrightarrow\) \(4x^4+4x^2+1\ge4x^2+1\)
\(\Leftrightarrow\) \(4x^4\ge0\) (đúng \(\forall x\)) \(\Rightarrow\) \(\dfrac{2x^2+1}{\sqrt{4x^2+1}}\ge1\) (đpcm)
a,
\(\dfrac{x^2+2}{\sqrt{x^2+1}}=\dfrac{\left(\sqrt{x^2+1}\right)^2+1}{\sqrt{x^2+1}}=\sqrt{x^2+1}+\dfrac{1}{\sqrt{x^2+1}}\ge2\)( Áp dụng bất đẳng thức AM - GM )
Vậy:
\(\dfrac{x^2+1}{\sqrt{x+1}}\ge2\)
Đẳng thức xảy ra khi và chỉ khi \(\sqrt{x^2+1}=\dfrac{1}{\sqrt{x^2+1}}\Rightarrow x=0\)