Cho \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)(với \(a,b,c\ne0;b\ne c\)) chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-d}\)
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T
2 tháng 2 2017
theo bài ra ta có:
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{b}{ab}+\frac{a}{ab}\right)\\ \Rightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\\ \Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
=> 2ab = c(a + b)
=> ab + ab = ca + cb
=> ab - cb = ca - ab
=> b( a - c ) = a( c - b )
=> \(\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
23 tháng 2 2019
\(\frac{1}{c}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})\)
\(\Rightarrow\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=(a+b)\cdot c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b(a-c)=a(c-b)\)
\(\frac{a}{c}=\frac{a-c}{c-b}(đpcm)\)
N
23 tháng 2 2019
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}.\left(\frac{a+b}{ab}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow ac+cb=2ab\Rightarrow ac-ab=-cb+ba\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
bn ghi sai đề kìa :v
13 tháng 7 2016
Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)
Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)
Áp dụng t/c bắc cầu ta dc : \(a=b=c\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)
13 tháng 7 2016
=>a2+b2=2ab
=>a2-2ab+b2=0
=>(a-b)2=0=>a=b
tương tự=>b=c
=>a=b=c
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)
24 tháng 9 2016
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{b}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=\left(1+1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
\(=3+\frac{a^2+b^2}{ab}+\frac{a^2+c^2}{ac}+\frac{b^2+c^2}{bc}\)
\(=3+\frac{a^2+b^2}{\frac{a^2+b^2}{2}}+\frac{a^2+c^2}{\frac{a^2+c^2}{2}}+\frac{b^2+c^2}{\frac{b^2+c^2}{2}}\)
\(=3+2+2+2=9\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ca+bc\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Trả lời :........................................................
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}......................\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Học sinh giỏi 6A