cho \(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{c}=0\)
tính: A= \(\dfrac{2bc}{a^2}+\dfrac{ca}{4b^2}+\dfrac{2ab}{c^2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 10 2021
\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)
Ta có:
\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 12 2020
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
LN
20 tháng 12 2019
cho mình hỏi bạn biết làm chưa nếu rồi thì giúp mình được không ạ mình ko biết làm
24 tháng 11 2021
\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
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\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
30 tháng 5 2023
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
HN
29 tháng 3 2017
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow ab+bc+ca=0\)
\(C=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
\(=\dfrac{a^2}{a^2+bc-ac-ab}+\dfrac{b^2}{b^2+ac-ba-bc}+\dfrac{c^2}{c^2+ab-ca-cb}\)
\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\left(\dfrac{a^2}{\left(a-b\right)\left(c-a\right)}+\dfrac{b^2}{\left(a-b\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\left(\dfrac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)=1\)
\(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{c}=0\Rightarrow2bc+ca+2ab=0\left(a;b;c\ne0\right)\)
\(\Rightarrow2bc=-a\left(2b+c\right);ca=-2b\left(a+c\right);2ab=-c\left(a+2b\right)\)
\(\Rightarrow A=-\dfrac{2b+c}{a}-\dfrac{a+c}{2b}-\dfrac{a+2b}{c}=\)
\(=\dfrac{-2bc\left(2b+c\right)-ac\left(a+c\right)-2ab\left(a+2b\right)}{2abc}=\)
\(=\dfrac{4b^2c+2bc^2+ac\left(a+c\right)+2a^2b+4ab^2}{-2abc}=\)
\(=\dfrac{2b\left(2bc+a^2+c^2+2ab\right)+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{2b\left[\left(a^2+c^2\right)+2bc+2ab\right]+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{2b\left[\left(a+c\right)^2-2ac+2bc+2ab\right]+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{2b\left[\left(a+c\right)^2-3ac+2bc+ac+2ab\right]+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{2b\left[\left(a+c\right)^2-3ac\right]+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{2b\left(a+c\right)^2-6abc+ac\left(a+c\right)}{-2abc}=\)
\(=\dfrac{\left(a+c\right)\left[2b\left(a+c\right)+ac\right]-6abc}{-2abc}=\)
\(=\dfrac{\left(a+c\right)\left(2ab+2bc+ac\right)-6abc}{-2abc}=\dfrac{-6abc}{-2abc}=3\)
Dạo này khi viết công thức cứ mất các dấu ngoặc là sao nhỉ