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Từ \(gt\Leftrightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\Leftrightarrow c\left(a+b\right)=2ab\Leftrightarrow ac+bc=ab+ab\)
\(\Leftrightarrow ac-ab=ab-bc\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{b-c}\) (đpcm)
Từ \(gt\Leftrightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\Leftrightarrow c\left(a+b\right)=2ab\Leftrightarrow ac+bc=ab+ab\)
\(\Leftrightarrow ac-ab=ab-bc\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{b-c}\)
\(\Rightarrowđpcm\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
=> \(\frac{2}{c}=\frac{a+b}{ab}\)
=> 2ab = ac + bc
=> ac + bc - 2ab = 0
=> (ac - ab) + (bc - ab) = 0
=> a(c - b) + b(c - a) = 0
=> a(c - b) = -b(c - a)
=> a(c - b) = b(a - c)
=> \(\frac{a}{b}=\frac{a-c}{c-b}\) (đpcm)
Ta có:
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{2}{c}\)
\(\Rightarrow\frac{b}{ab}+\frac{a}{ab}=\frac{2}{c}\)
\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{2}{c}\)
\(\Rightarrow2ab=\left(a+b\right).c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right).\)
Chúc bạn học tốt!
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Leftrightarrow ca+cb=2ab\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Leftrightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ca+bc\)
\(ab-cb=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)