Tìm ba số dương a,b,c biết
\(\frac{\sqrt{ab}-1}{3}=\frac{\sqrt{bc}-3}{9}=\frac{\sqrt{ca}-5}{-6}và\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=11\)
Help me đi các bn eh !
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2019
\(\frac{\sqrt{ab}-1}{3}=\frac{\sqrt{bc}-3}{9}=\frac{\sqrt{ac}-5}{-6}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}-9}{6}=\frac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{\sqrt{ab}-1}{3}=\frac{1}{3}\\\frac{\sqrt{bc}-3}{9}=\frac{1}{3}\\\frac{\sqrt{ac}-5}{-6}=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{ab}=2\\\sqrt{bc}=6\\\sqrt{ac}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=4\\bc=36\\ac=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=9a\\ac=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=9\end{matrix}\right.\)
22 tháng 4 2020
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
AH
Akai Haruma
Giáo viên
31 tháng 3 2018
Lời giải:
Đặt \(\frac{\sqrt{ab}-1}{3}=\frac{\sqrt{bc}-3}{9}=\frac{\sqrt{ca}-5}{-6}=t\)
\(\Rightarrow \left\{\begin{matrix} \sqrt{ab}=3t+1\\ \sqrt{bc}=9t+3\\ \sqrt{ca}=5-6t\end{matrix}\right.\)
\(\Rightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6t+9\)
\(\Leftrightarrow 11=6t+9\Leftrightarrow t=\frac{1}{3}\)
Khi đó : \(\left\{\begin{matrix} \sqrt{ab}=2\\ \sqrt{bc}=6\\ \sqrt{ac}=3\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} ab=4\\ bc=36\\ ac=9\end{matrix}\right.\Rightarrow abc=\sqrt{4.36.9}=36\)
\(\Rightarrow \left\{\begin{matrix} c=\frac{abc}{ab}=9\\ a=\frac{abc}{bc}=1\\ b=\frac{abc}{ac}=4\end{matrix}\right.\)
Vậy....
YY
21 tháng 5 2018
\(\frac{\sqrt{ab}}{c+2\sqrt{ab}}=\frac{1}{2}\left(\frac{x+2\sqrt{xy}-z}{z+2\sqrt{xy}}\right)=\frac{1}{2}\left(1-\frac{z}{z+2\sqrt{xy}}\right)\le\frac{1}{2}\left(1-\frac{z}{x+y+z}\right)\)
Tương tự \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}\le\frac{1}{2}\left(1-\frac{x}{x+y+z}\right)\);\(\frac{\sqrt{xz}}{y+2\sqrt{xz}}\le\frac{1}{2}\left(1-\frac{y}{x+y+z}\right)\)
Cộng vế theo vế ta được \(\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\le\frac{1}{2}\left(3-1\right)=1\)
YY
25 tháng 5 2018
\(c+ab=\left(a+b+c\right)c+ab=ac+cb+c^2+ab=\left(a+c\right)\left(b+c\right)\)
Tương tự : \(a+bc=\left(a+b\right)\left(a+c\right);c+ab=\left(c+a\right)\left(c+b\right)\)
\(P=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)
áp dụng bất đẳng tức cauchy :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
cộng vế theo vế
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}+\frac{a}{b+a}\right)\)
\(\Leftrightarrow P\le\frac{1}{2}\left(\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{1}{2}\cdot3=\frac{3}{2}\)
dấu "=" xảy ra khi a=b=c=1/3
24 tháng 8 2020
Có a+b+c=1 => c=(a+b+c).c=ac+bc+c2
\(\Rightarrow c+ab=ac+bc+c^2+ab=a\left(b+c\right)+c\left(b+c\right)=\left(b+c\right)\left(a+c\right)\)
\(\Rightarrow\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{\frac{a}{c+b}+\frac{b}{c+b}}{2}\)
Tương tự ta có \(\hept{\begin{cases}a+bc=\left(a+b\right)\left(a+c\right)\\b+ac=\left(b+a\right)\left(b+c\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{b+ca}}=\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\end{cases}}}\)
\(\Rightarrow P\le\frac{\frac{b}{a+b}+\frac{c}{c+a}+\frac{c}{b+c}+\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{c+b}}{2}\)\(=\frac{\frac{a+c}{a+c}+\frac{c+b}{c+b}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)