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ta có : \(2x^2-5x-7=2\left(x^2-2.\frac{5}{4}+\frac{25}{16}\right)-\frac{81}{8}\)
\(=2\left(x-\frac{5}{4}\right)^2-\frac{81}{8}\ge-\frac{81}{8}\)
vậy min =\(-\frac{81}{8}\Leftrightarrow x=\frac{5}{4}\)
ta có: \(8x^2-20x+5=8\left(x^2-2.\frac{5}{4}x+\frac{25}{16}\right)-\frac{15}{2}\)
\(=8\left(x-\frac{5}{4}\right)^2-\frac{15}{2}\ge\frac{-15}{2}\)
vậy min=\(\frac{-15}{2}\Leftrightarrow x=\frac{5}{4}\)
đkxđ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
A = \(\frac{1}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+1}\)
= \(\frac{2\sqrt{a}}{a-1}\)
ta có \(6-\sqrt{5}>6-\sqrt{9}=3=\sqrt{16}-\sqrt{1}>4-\sqrt{3}\)
ta có: \(\left(4+\sqrt{15}\right)\sqrt{2\left(\sqrt{5}-\sqrt{3}\right)}\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{3}.\sqrt{5}}\)
\(=\frac{\left(8+2\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}\)
\(=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{5}-\sqrt{3}\right)^2}{2}\)
\(=\frac{\left(5-3\right)^2}{2}=2\)
đkxđ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\\\sqrt{x}\left(3-\sqrt{x}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne4\\x\ne9\end{matrix}\right.\)
a) \(Q=\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\frac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)
b) Để Q=2 thì \(2(\sqrt{x}-3)=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=8\)
\(\Leftrightarrow x=64\)
c) Vì \(\sqrt{x}+2\ge2>0\) nên để Q<0 thì \(\sqrt{x}< 3\) hay x<9
xét \(_{\Delta}\)AOB vuông tại O có:
\(\frac{1}{h^2}=\frac{1}{OA^2}+\frac{1}{OB^2}\) hay \(\frac{1}{h^2}=\frac{1}{\left(\frac{m}{2}\right)^2}+\frac{1}{\left(\frac{n}{2}\right)^2}\)
\(\Leftrightarrow\frac{1}{h^2}=4\left(\frac{1}{m^2}+\frac{1}{n^2}\right)\)
=> đpcm
A B C D O h m n
đkxđ: x>0
\(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{8\sqrt{x}+8}{x+2\sqrt{x}}-\frac{\sqrt[]{x}+2}{\sqrt{x}}\right):\frac{x+2\sqrt{x}+5}{x+2\sqrt{x}}\)
= \(\left[\frac{\sqrt{x}.\sqrt{x}+8\sqrt{x}+8-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\sqrt{x}}\right]:\frac{x+2\sqrt{x}+5}{x+2\sqrt{x}}\)
=\(\frac{4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\sqrt{x}}.\frac{\left(\sqrt{x}+2\right)\sqrt{x}}{x+2\sqrt{x}+5}\)
=\(\frac{4\sqrt{x}+4}{x+2\sqrt{x}+5}\)
áp dụng bđt cô si thôi bạn bđt cho 2 số a,b không âm : a+b\(\ge2\sqrt{ab}\) a. \(P=2\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2\sqrt{x}.\frac{1}{\sqrt{x}}}=2\sqrt{2}\)
vậy Min P=\(2\sqrt{2}\) \(\Leftrightarrow2\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\)
b. \(P=4\sqrt{x}-1+\frac{3}{\sqrt{x}}\)
\(\ge2\sqrt{4\sqrt{x}.\frac{3}{\sqrt{x}}}-1=-1+4\sqrt{3}\)
vậy Min P \(=-1+4\sqrt{3}\) \(\Leftrightarrow4\sqrt{x}=\frac{3}{\sqrt{x}}\Leftrightarrow x=\frac{3}{4}\)