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Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Leftrightarrow ca+cb=2ab\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Leftrightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{b+a}{2ab}\right)\)
\(\frac{1}{c}=\frac{b+a}{2ab}\)
suy ra \(2ab=c\left(b+a\right)\)
\(2ab=cb+ca\)
suy ra \(ab+ab=cb+ca\)
suy a \(ab-cb=ca-ab\)
suy ra \(b\left(a-c\right)=a\left(c-b\right)\)
suy ra \(\frac{a}{b}=\frac{a-c}{c-b}\left(Đpcm\right)\)
\(\frac{a-c}{c-b}=\frac{a}{b}\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow1.2ab=c.\left(a+b\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}=\frac{a}{2ab}+\frac{b}{2ab}=\frac{1}{2b}+\frac{1}{2a}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\text{Vậy }\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
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 \(gt\Leftrightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\Leftrightarrow c\left(a+b\right)=2ab\Leftrightarrow ac+bc=ab+ab\)
\(\Leftrightarrow ac-ab=ab-bc\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{b-c}\)
\(\Rightarrowđpcm\)
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)
Xét a+b+c=0
\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
Xét a+b+c\(\ne0\)
\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{2c}{b}\cdot\frac{2a}{c}\cdot\frac{2b}{a}=8\)
Giải:
+) Xét a + b + c = 0
\(\Rightarrow-a=b+c\)
\(\Rightarrow-b=a+c\)
\(\Rightarrow-c=a+b\)
Ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{-c}{c}=\frac{-a}{a}=\frac{-b}{b}=-1\)
Lại có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=-1\)
+) Xét \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Ta có:
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)
Vậy M = -1 hoặc M = 8
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\Rightarrow\frac{1}{c}=\frac{2a+2b}{4ab}\Rightarrow4ab=2ac+2bc\)
\(\frac{a}{b}=\frac{a-c}{c-b}\Rightarrow ac-ab=ab-bc\Rightarrow2ab=ac+bc\Rightarrow4ab=2ac+2bc\) (chứng minh trên)
Vậy \(\frac{a}{b}=\frac{a-c}{c-b}\)
Bài này hình như mk có làm rooid nhưng chẳng nhớ