Bài 1: CMR:
a, (4+\(\sqrt{3}\)). (4-\(\sqrt{3}\))=13
b, \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=2\)
c, \(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=4\)
d, \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=a-b\)(a>0, b>0, a≠b)
Bài 2: CMR:
a, \(\sqrt{a}+\frac{\sqrt{1}}{\sqrt{a}}\ge2\left(a>0\right)\)
b, a+b+\(\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\left(a,b>0\right)\)
c, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xyz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\left(x,y,z>0\right)\)
d, \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=-8\sqrt{3}\)
e, \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)=a-b(a>0, b>0, a≠b)
Bài 3: Tìm Min hoặc Max(nếu có):
a, \(\sqrt{x^2+9}\)
b, \(\frac{2}{\sqrt{x^2+1}}\)
c, 1-\(\sqrt{5+2x-x^2}\)
2/
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)
Giả sử các biểu thức đều xác định
1/ \(\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)=4^2-\left(\sqrt{3}\right)^2=16-3=13\)
\(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1-\left(\sqrt{7}-1\right)=2\)
\(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{2-\sqrt{3}}{4-3}+\frac{2+\sqrt{3}}{4-3}=4\)
\(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\left(\sqrt{a}-\sqrt{b}\right)=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
a/ Do \(x^2\ge0\Rightarrow x^2+9\ge9\Rightarrow A=\sqrt{x^2+9}\ge\sqrt{9}=3\)
\(\Rightarrow A_{min}=3\) khi \(x=0\) , \(A_{max}\) ko tồn tại
b/ Do \(\left\{{}\begin{matrix}2>0\\\sqrt{x^2+1}\ge1\end{matrix}\right.\) \(\Rightarrow A=\frac{2}{\sqrt{x^2+1}}\le\frac{2}{1}=2\)
\(\Rightarrow A_{max}=2\) khi \(x=0\), \(A_{min}\) không tồn tại
c/ Ta có \(\sqrt{5+2x-x^2}=\sqrt{6-\left(x^2-2x+1\right)}=\sqrt{6-\left(x-1\right)^2}\)
Biểu thức xác định khi
\(6-\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2\le6\Rightarrow1-\sqrt{6}\le x\le1+\sqrt{6}\)
Khi đó: \(\left\{{}\begin{matrix}\sqrt{5+2x-x^2}\ge0\\\sqrt{5+2x-x^2}=\sqrt{6-\left(x-1\right)^2}\le\sqrt{6}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A=1-\sqrt{5+2x-x^2}\le1\\A=1-\sqrt{5+2x-x^2}\ge1-\sqrt{6}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A_{max}=1\Leftrightarrow x=\left[{}\begin{matrix}1-\sqrt{6}\\x=1+\sqrt{6}\end{matrix}\right.\\A_{min}=1-\sqrt{6}\Leftrightarrow x=1\end{matrix}\right.\)
