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19 tháng 11 2016
1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
19 tháng 11 2016
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
2 tháng 9 2019
\(2,\)
\(a,\sqrt{x^2-4x+3}=3\)
\(\Rightarrow x^2-4x+3=9\)
\(\Rightarrow x^2-4x-6=0\)
\(\Rightarrow\left(x-2\right)^2=10\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{10}\\x-2=-\sqrt{10}\end{cases}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{cases}}}\)
BT
26 tháng 5 2018
a/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
Và: \(x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=> \(P=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}\)
=> \(P=\frac{2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)
b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\) => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)
=> \(P=-\left(2+\sqrt{3}\right)\)
c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)
<=> \(\frac{1}{\sqrt{x}-1}=-2\)
<=> \(1=-2\sqrt{x}+2\)
<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)
\(x+2=3\sqrt{1-x^2}+\sqrt{1+x}\)
\(ĐKXĐ:-1\le x\le1\)
\(x+2=3\sqrt{1-x}\sqrt{1+x}+\sqrt{1+x}\)
\(\left(3\sqrt{1-x}\sqrt{1+x}-\frac{3}{2}\right)+\left(\sqrt{1+x}-x-\frac{1}{2}\right)=0\)
\(\frac{9\left(1-x\right)\left(1+x\right)-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-\left(x+\frac{1}{2}\right)^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9-9x^2-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-x^2-x-\frac{1}{4}}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{\frac{27}{4}-9x^2}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9\left(\frac{3}{4}-x^2\right)}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\left(\frac{3}{4}-x^2\right)\left(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}\right)=0\)
\(\orbr{\begin{cases}\frac{3}{4}-x^2=0\\\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}=0\left(KTM\right)\end{cases}< =>x=\frac{\sqrt{3}}{2}\left(TM\right)}\)
\(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}>0\)nên pt ktm