Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)
\(\Rightarrow\frac{b}{ab}+\frac{a}{ba}=\frac{2}{c}\)
\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)
\(\Rightarrow\left(b+a\right)c=2ab\)
\(\Rightarrow cb+ca=ab+ab\)
\(\Rightarrow ca-ab=ab-cb\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a-c}{c-b}=\frac{a}{b}\)
\(\frac{1}{c}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})\)
\(\Rightarrow\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=(a+b)\cdot c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b(a-c)=a(c-b)\)
\(\frac{a}{c}=\frac{a-c}{c-b}(đpcm)\)
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}.\left(\frac{a+b}{ab}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow ac+cb=2ab\Rightarrow ac-ab=-cb+ba\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
bn ghi sai đề kìa :v
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
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(2\right)\)
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
Áp dụng t/c dãy tỉ số = nhau
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\frac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\)
Tương tự \(b+c=2a;;c+a=2b\)
\(\Rightarrow D=\left(\frac{a+b}{a}\right)\left(\frac{b+c}{b}\right)\left(\frac{c+a}{c}\right)=\left(\frac{2c}{a}\right)\left(\frac{2a}{b}\right)\left(\frac{2b}{c}\right)=8\)
Theo đề ta có :
\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{a+c-b}{b}+2\)
\(\Rightarrow\frac{a+b-c+2c}{c}=\frac{b+c-a+2a}{a}=\frac{a+c-b+2b}{b}\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow\left(a+b+c\right).\frac{1}{c}=\left(a+b+c\right)\frac{1}{c}=\left(a+b+c\right)\frac{1}{b}\)
(vì \(a\ne b\ne c\ne0\) \(\frac{\Rightarrow1}{a}\ne\frac{1}{b}\ne\frac{1}{c}\ne0\) \(\Rightarrow a+b+c=0\))
* a+b+c=0
=>a+b=-c ; b+c=-a ; a+c =-b
\(D=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
\(=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c.-a.-b}{a.b.c}=\frac{-1.\left(a.b.c\right)}{a.b.c}=-1\)
Vậy : D=-1