Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
=> \(\frac{2}{c}=\frac{a+b}{ab}\)
=> 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 \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
áp dụng dãy tỉ số bằng nhau ta có
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Áp dụng tỉ dãy số bằng nhau. Ta có:
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Leftrightarrow\frac{1+1+1}{a+b+c}=1\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{a}{b}\Leftrightarrow1-1\Leftrightarrow0\)
\(\Rightarrow PT=\frac{a-c}{c-b}=\frac{\left(a-c\right)^0}{\left(c-b\right)^0}=0\)
Vậy dấu = xảy ra khi a - c = a , c - b = b
Ta có ĐPCM
Ps: Chả biết đúng hay không nữa
như này mới đúng nè
ta có\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)
\(\Rightarrow\frac{b}{ab}+\frac{a}{ba}=\frac{2}{c}\)
\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)
\(\Rightarrow\left(b+a\right)c=2ab\)
\(\Rightarrow cb+ca=ab+ab\)
\(\Rightarrow ca-ab=ab-cb\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a-c}{c-b}=\frac{a}{b}\)
\(\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=\frac{a+b+a+c+b+c}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
- Nếu \(a+b+c=0\)
\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
- Nếu \(a=b=c\Rightarrow M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(2\right)\)
Mà \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(1\right)\)
Nhận thấy ( 1 )=( 2 ) => đpcm
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow c=\frac{1}{\frac{1}{2a}+\frac{1}{2b}}=\frac{1}{\frac{2\left(a+b\right)}{4ab}}=\frac{4ab}{2\left(a+b\right)}=\frac{2ab}{a+b}\)
\(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{a\left(1-\frac{2b}{a+b}\right)}{b\left(\frac{2a}{a+b}-1\right)}=\frac{a\left(\frac{a-b}{a+b}\right)}{b\left(\frac{a-b}{a+b}\right)}=\frac{a}{b}\)
\(\RightarrowĐPCM\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\Rightarrow2ab=\left(a+b\right).c\)
\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Giải
Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{c}\div\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{1}{c}\times\frac{2}{1}=\frac{b}{ab}+\frac{a}{ab}\)
\(\Leftrightarrow\frac{2}{c}=\frac{b+a}{ab}\)
\(\Leftrightarrow2ab=c\left(b+a\right)\)
\(\Leftrightarrow ab+ab=bc+ac\)
\(\Leftrightarrow ac-ab=bc-ab\)
\(\Leftrightarrow a\left(c-b\right)=b\left(c-a\right)\)
Từ đẳng thức trên , ta áp dụng tính chất của tỉ lệ thức :
\(\frac{a}{b}=\frac{a-c}{c-b}\)