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10 tháng 8 2019

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\)

\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

19 tháng 8 2019

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\) ( do \(a+b+c=0\) )

\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)

\(=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) ( đpcm )

21 tháng 12 2019

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)

\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)

\(\Rightarrow2ab=c.\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\)

\(\Rightarrow ab-bc=ac-ab\)

\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)

6 tháng 1 2017

ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)

\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)

\(=\frac{2}{c}=\frac{b+a}{ab}\)

= \(c\left(b+a\right)=ab\times2\)

= cb +ca = ab+ab

= ab - cb = ac-ab

\(=b\left(a-c\right)=a\left(c-b\right)\)

= \(\frac{a}{b}=\frac{a-c}{c-b}\)

6 tháng 1 2017

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)

\(\frac{1}{c}=\frac{a+b}{2ab}\)

\(2ab=c\left(a+b\right)\)

\(ab+ab=ac+bc\)

\(ab-bc=ac-ab\)

\(b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)

6 tháng 10 2017

Áp dụng tỉ dãy số bằng nhau. Ta có:

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Leftrightarrow\frac{1+1+1}{a+b+c}=1\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{a}{b}\Leftrightarrow1-1\Leftrightarrow0\)

\(\Rightarrow PT=\frac{a-c}{c-b}=\frac{\left(a-c\right)^0}{\left(c-b\right)^0}=0\)

Vậy dấu = xảy ra khi a - c = a               , c - b = b

Ta có ĐPCM

Ps: Chả biết đúng hay không nữa

29 tháng 12 2020

như này mới đúng nè 

ta có\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)

\(\Rightarrow\frac{b}{ab}+\frac{a}{ba}=\frac{2}{c}\)

\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)

\(\Rightarrow\left(b+a\right)c=2ab\)

\(\Rightarrow cb+ca=ab+ab\)

\(\Rightarrow ca-ab=ab-cb\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a-c}{c-b}=\frac{a}{b}\)

20 tháng 6 2017

\(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ca+b}{\left(c+a\right)^2}\)

\(=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)

\(=\frac{\left(c+a\right)\left(c+b\right)}{\left(a+b\right)^2}.\frac{\left(a+b\right)\left(a+c\right)}{\left(b+c\right)^2}.\frac{\left(b+a\right)\left(b+c\right)}{\left(c+a\right)^2}=1\)