CHO \(\hept{\begin{cases}a+bkhác0\\a,bkhác0\end{cases}}\)
CMR
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=trituyetdoi\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)\)
K
Khách
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Những câu hỏi liên quan
9 tháng 9 2017
1,
\(A=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)
\(\ge1+1+2\sqrt{\frac{a}{b}.\frac{b}{a}}+a+b+\frac{a+b}{ab}=4+a+b+\frac{4\left(a+b\right)}{\left(a+b\right)^2}=4+a+b+\frac{4}{a+b}\)
lại có \(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a+b\le\sqrt{2}\)
\(4+a+b+\frac{4}{a+b}=4+\left(a+b+\frac{2}{a+b}\right)+\frac{2}{a+b}\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
\(\Rightarrow A\ge4+3\sqrt{2}\)
9 tháng 9 2017
câu 2
ta có:\(\left(2b^2+a^2\right)\left(2+1\right)\ge\left(2b+a\right)^2\Rightarrow3c\ge a+2b\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\left(Q.E.D\right)\)
12 tháng 9 2016
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2}{ab}}\)
\(=\sqrt{\left(\frac{a+b}{ab}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2\left(a+b\right)}{ab}.\frac{1}{a+b}}\)
\(=\sqrt{\left(\frac{a+b}{ab}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
19 tháng 7 2021
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
FM
18 tháng 9 2018
ĐKXĐ: \(a\ne1;b\ne2;c\ne3\)
Đặt \(\frac{1}{a-1}=x;\frac{1}{b-2}=y;\frac{1}{c-3}=z\). Khi đó hệ phương trình đã cho tương đương:
\(\hept{\begin{cases}x+y+z=1\\x^2-2yz=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1-y-z\\\left(1-y-z\right)^2-2yz=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1-y-z\\y^2+z^2+1+2yz-2y-2z-2yz=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1-y-z\\y^2+z^2-2y-2z+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1-y-z\\\left(y-1\right)^2+\left(z-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\\z=1\end{cases}}\).
VT
26 tháng 6 2018
Ta có\(ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=1\)
Thay 1=ab+bc+ca vào, ta có
\(a\sqrt{\frac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}=a\sqrt{\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}{\left(a+b\right)\left(a+c\right)}}=a\left(b+c\right)\)
Tương tự rồi cộng lại, ta có
A=2(ab+bc+ca)=2
^_^
XO
25 tháng 8 2020
Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)
Khi đó Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020
= 1 + 1 + 1 = 3
Vậy P = 3
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2}{ab}}\)
\(=\sqrt{\left(\frac{a+b}{ab}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2\left(a+b\right)}{ab}.\frac{1}{a+b}}\)
\(=\sqrt{\left(\frac{a+b}{ab}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
thanks kb nhe