Ta có :

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{ab-bc}{\left(a+b\right)-\left(b+c\right)}=\frac{bc-ca}{\left(b+c\right)-\left(c+a\right)}=\frac{ab-ca}{\left(a+b\right)-\left(c+a\right)}\)

\(\Rightarrow a=b=c\)

\(\Rightarrow Q=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=1\)

Ta có: \(\frac{ab + ac}{2} = \frac{bc + ba}{3} = \frac{ca + cb}{4}\)

=>\(\frac{a(b + c)}{2} = \frac{b(c + a)}{3} = \frac{c(a + b)}{4}\)

=>\(\frac{a(b + c)}{2abc} = \frac{b(c + a)}{3abc} = \frac{c(a + b)}{4abc}\)

=>\(\frac{b + c}{2bc} = \frac{c + a}{3ca} = \frac{a + b}{4ab}\)

=>\(\frac{b}{2bc} + \frac{c}{2bc} = \frac{c}{3ca} + \frac{a}{3ca} = \frac{a}{4ab} + \frac{b}{4ab}\)

=>\(\frac{1}{2c}+\frac{1}{2b}=\frac{1}{3a}+\frac{1}{3c}=\frac{1}{4b}+\frac{1}{4a}\)

=>\(12 \cdot \left(\frac{1}{2c} + \frac{1}{2b}\right) = 12 \cdot \left(\frac{1}{3a} + \frac{1}{3c}\right) = 12 \cdot \left(\frac{1}{4b} + \frac{1}{4a}\right)\)

=>\(\frac{6}{c}+\frac{6}{b}=\frac{4}{a}+\frac{4}{c}=\frac{3}{b}+\frac{3}{a}\quad(*)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{4}{a} + \frac{4}{c}\)

=>\(\frac{6}{c}-\frac{4}{c}+\frac{6}{b}=\frac{4}{a}\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{a}\quad(1)\)

Ta có: \(\frac{4}{a} + \frac{4}{c} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{4}{a}-\frac{3}{a}+\frac{4}{c}=\frac{3}{b}\)

=>\(\frac{1}{a}+\frac{4}{c}=\frac{3}{b}\implies\frac{3}{b}-\frac{1}{a}=\frac{4}{c}\quad(2)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{6}{b}-\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\)

=>\(\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\implies\frac{1}{b}+\frac{2}{c}=\frac{1}{a}\quad(3)\)

Từ (1),(3) suy ra \(\frac{2}{c} + \frac{6}{b} = 4 \left(\frac{1}{b} + \frac{2}{c}\right)\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{b}+\frac{8}{c}\)

=>\(\frac{-6}{c}=\frac{-2}{b}\)

=>\(\frac{6}{c}=\frac{2}{b}\)

=>\(\frac{c}{6}=\frac{b}{2}\)

=>\(\frac{c}{15}=\frac{b}{5}\) (4)

=>\(\frac{1}{b}=\frac{3}{c}\)

=>\(\frac{1}{a} = \frac{3}{c} + \frac{2}{c} = \frac{5}{c}\)

=>\(\frac{a}{1}=\frac{c}{5}\)

=>\(\frac{a}{3}=\frac{c}{15}\) (5)

Từ (4),(5) suy ra \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\) (ĐPCM)

a,Theo gt, ta có :\(a.\left(a-b\right)-b.\left(a-b\right)=64\Rightarrow\left(a-b\right)^2=64\Rightarrow\)\(\Rightarrow a-b=8\left(1\right)\)

Lại có:\(a.\left(a-b\right)+b.\left(a-b\right)=-16\Rightarrow\left(a+b\right).\left(a-b\right)=-16.\left(2\right)\)\(Thay:a-b=8\)vào \(\left(2\right)\) ta được:

\(\left(a+b\right).8=-16\Rightarrow a+b=-2\left(3\right)\)

Từ \(\left(1\right)\)và \(\left(3\right)\)\(\Rightarrow\hept{\begin{cases}a=3\\b=-5\end{cases}}\)

b, Theo gt, ta có :\(a.b.b.c.c.a=\frac{1}{16}\Rightarrow\left(a.b.c\right)^2=\frac{1}{16}\Rightarrow a.b.c=\frac{1}{4}\)\(\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=-\frac{2}{3}\\c=-\frac{3}{4}\end{cases}}\)

bạn dùng TC dãy tỉ số bằng nhau đi

cộng vào là ra kết quả ngay mà

\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{bc}{b+c}\Rightarrow\frac{abc}{c\left(a+b\right)}=\frac{abc}{b\left(a+c\right)}=\frac{abc}{a\left(b+c\right)}\)

\(\Rightarrow c\left(a+b\right)=b\left(a+c\right)\Leftrightarrow ac+bc=ab+bc\Rightarrow ac=ab\Rightarrow c=b\) (1)

\(\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Leftrightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow b=a\) (2)

\(\Rightarrow c\left(a+b\right)=a\left(b+c\right)\Leftrightarrow ac+bc=ab+ac\Rightarrow bc=ab\Rightarrow c=a\) (3)

Từ (1) ; (2) ; (3) => \(a=b=c\) (ĐPCM)

Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) => \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\) => \(\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)

=> \(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\) => \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\) => a = b = c

Vậy B = \(\frac{a.a^2+b.b^2+c.c^2}{a^3+b^3+c^3}=\frac{a^3+b^3+c^3}{a^3+b^3+c^3}=1\)