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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)
\(\frac{2}{a}=\frac{1}{b}+\frac{1}{c}=\frac{b+c}{bc}\)
=> 2bc = a.(b + c)
=> bc + bc = ab + ac
=> bc - ac = ab - bc
=> c.(b - a) = b.(a - c)
\(\Rightarrow\frac{b-a}{a-c}=\frac{b}{c}\left(đpcm\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ca+bc\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
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
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
