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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 )
=> \(\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ca+bc\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
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
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
=> \(\frac{2}{c}=\frac{a+b}{ab}\)
=> 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)
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: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{b+a}{ab}\)
= \(c\left(b+a\right)=ab\times2\)
= cb +ca = ab+ab
= ab - cb = ac-ab
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)