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a)
\(\left(\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}\right)^2\)
\(=a+\sqrt{b}\ne2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}+a-\sqrt{b}\)
\(=2a\ne2\sqrt{a^2-b}=2\left(a\ne\sqrt{a^2}-b\right)\)
\(\Rightarrow\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\ne\sqrt{a^2}-b\right)}\)
\(\Rightarrowđpcm\)
b)
\(\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}\ne}\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2\)
\(=\frac{a+\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a+\sqrt{a^2-b}}{2}.\frac{a-\sqrt{a^2-b}}{2}}+\frac{a-\sqrt{a^2-b}}{2}\)
\(=\frac{a}{2}+\frac{\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a^2-a^2+b}{2.2}}+\frac{a}{2}-\frac{\sqrt{a^2-b}}{2}\)
\(=a\ne2\frac{\sqrt{b}}{2}=a\ne\sqrt{b}\)
\(\Rightarrow\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\ne\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{a\ne\sqrt{b}}\)
\(\Rightarrowđpcm\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
a) \(a+b-2\sqrt{ab}\ge0\)
<=> \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge0\) (luôn đúng )
=> đpcm
b) \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\Leftrightarrow\sqrt{\dfrac{a+b}{2}^2}\ge\left(\dfrac{\sqrt{a}+\sqrt{b}}{2}\right)^2\)
<=> \(\dfrac{a+b}{2}\ge\dfrac{a+b+2\sqrt{ab}}{4}\)
<=> \(\dfrac{2a+2b}{4}\ge\dfrac{a+b+2\sqrt{ab}}{4}\Leftrightarrow2a+2b\ge a+b+2\sqrt{ab}\)
<=> \(2a+2b-a-b-2\sqrt{ab}\ge0\)
<=> \(a-2\sqrt{ab}+b\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
=> đpcm
Với b\(\ge\)0, a\(\ge\)\(\sqrt{b}\) ta bình phương 2 vế lên có:
\(\sqrt{a\pm \sqrt{b}}^2\)=\((\sqrt{\dfrac{\sqrt{a+\sqrt{a^2-b}}}{2}}\)\pm \(\sqrt{\dfrac{\sqrt{a-\sqrt{a^2-b}}}{2}})^2\)
Xét vế trái ta có:
\(\sqrt{(a\pm \sqrt{b})^2}\)=\(a\pm \sqrt{b})