Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)
\(a,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}=\dfrac{a^2+c^2}{b^2+c^2}\left(1\right)\)
Mà \(\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\Leftrightarrow\dfrac{a}{b}=\dfrac{c^2}{b^2}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\tođpcm\)
\(b,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\)
\(\Leftrightarrow\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b-a\right)\left(b+a\right)}{a^2+ab}=\dfrac{\left(b-a\right)\left(b+a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\left(đpcm\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow ac=b^2\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)
Đặt \(S=a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)\)
Từ giả thiết: \(a+b+c=0\Rightarrow b+c=-a;c+a=-b;a+b=-c.\)
Thay vào biểu thức S, ta có:
\(S=a^2.\left(-a\right)+b^2.\left(-b\right)+c^2.\left(-c\right)=-a^3-b^3-c^3\)
\(S=-\left(a^3+b^3+c^3\right)=-\left[\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]\)
\(S=-\left[0-3\left(-c\right).\left(-a\right).\left(-b\right)\right]\)(Do a+b+c=0 và a+b=-c; b+c=-a; a+b=-c)
\(S=-\left[-3.\left(-abc\right)\right]=-\left(3abc\right)\)
Thay \(abc=-15\)vào biểu thức S: \(S=-\left[3.\left(-15\right)\right]=-\left(-45\right)=45.\)
ĐS: \(S=45.\)