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a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(A=\left|1-x\right|-1=1-x-1=-x\)
\(B=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\sqrt{x}-3\)
\(C=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(D=\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x=\left[{}\begin{matrix}-1\left(x\ge1\right)\\1-2x\left(x< 1\right)\end{matrix}\right.\)
\(D=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(D=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3\sqrt{x}+1-\left(3\sqrt{x}-2\right)}\)
\(D=\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3\sqrt{x}+1-3\sqrt{x}+2}\)
\(D=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}\)
\(D=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)