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 )

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

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)

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

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

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

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

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

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

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

Trả lời :........................................................

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

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A

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

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

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

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

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

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

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

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

1. 

Ta có : \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}\)

\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)

Áp dụng tính chất của dãy tỉ số bằng hau ta có :

 \(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)

\(=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)

\(\Rightarrow\hept{\begin{cases}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{x}{3c}\left(đpcm\right)\)

Chúc bạn học tốt !!!

1. Sửa lại dòng cuối 

\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)

Bài 2:

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

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

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

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

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

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

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