chờ a,b,c khác 0 và
\(\frac{2}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=\(\frac{2}{a+2b+2c}\)
CMR : (a+2b)(b+c)(2c+a)=0
Những câu hỏi liên quan
Cho a, b, c \(\ne\)0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\). Tính : \(E=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-a^2c^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}.\)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Cho a,b,c khác 0 và thỏa mãn: \(\frac{2ab+1}{2b}=\frac{2bc+1}{c}=\frac{ac+1}{a}\). CMR: a=2b=c hoặc \(4a^2b^2c^2=1\)
Cho a,b,c > 0 . CMR : \(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\)≥\(1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(C=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(c=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC
Cho a,b,c > 0.CMR:
a, \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
b, \(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
a) Dùng (a+b)2≥4ab
Chia hai vế cho a+b ( vì ab khác 0)
Ta có a+b≥\(\frac{4ab}{a+b}\) (Chuyển ab sang a+b) ta có
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)
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Cho a,b,c>0 CMR
\(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
cho a,b,c> 0 . Cmr:
\(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
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Câu hỏi của khoimzx - Toán lớp 9 | Học trực tuyến
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Cho a,b,c > 0. CMR:
\(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
\(\frac{a}{b+2c}+\frac{a}{b+2a}\ge\frac{4a}{2a+2b+2c}=\frac{2a}{a+b+c}\)
Tương tự: \(\frac{b}{c+2a}+\frac{b}{c+2b}\ge\frac{2b}{a+b+c}\) ; \(\frac{c}{a+2b}+\frac{c}{a+2c}\ge\frac{2c}{a+b+c}\)
Cộng vế với vế:
\(\Rightarrow\frac{1}{2}.VT+\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}\ge2\)
\(\Leftrightarrow VT+\frac{2a}{b+2a}+\frac{2b}{c+2b}+\frac{2c}{a+2c}\ge4\)
\(\Leftrightarrow VT+\left(1-\frac{b}{b+2a}\right)+\left(1-\frac{c}{c+2b}\right)+\left(1-\frac{a}{a+2c}\right)\ge4\)
\(\Leftrightarrow VT\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
Dấu "=" xảy ra khi \(a=b=c\)
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