Tham khảo: Câu hỏi của Đậu Đình Kiên

Áp dụng tính chất hãy tỉ số bằng nhau ta có:

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

\(\Rightarrow a+b=2c;b+c=2a;a+c=2b\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{b}{a}=\frac{a}{c}=\frac{c}{b}=1\)

\(\Rightarrow B=2.2.2=8\)

ta có: \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a-a+a+b+b-b-c+c+c}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)

 nếu a+b+c =0

=> a =0-b-c => a = -(b+c)

b     = 0-a-c => b = -(a+c)

c      = 0-a-b => c = -(a+b)

thay vào \(B=\left(1+\frac{-\left(a+c\right)}{a}\right).\left(1+\frac{-\left(b+c\right)}{c}\right).\left(1+\frac{-\left(a+b\right)}{b}\right)\)

\(B=\left(\frac{a-\left(a+c\right)}{a}\right).\left(\frac{c-\left(b-c\right)}{c}\right).\left(\frac{b-\left(a+b\right)}{b}\right)\)

\(B=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}\)

\(B=-1\)

 nếu a+b+c khác 0

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

=> \(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)\)

\(B=\left(1+1\right).\left(1+1\right).\left(1+1\right)\)

\(B=2.2.2\)

\(B=8\)

KL: B= -1 hoặc B=8

Chúc bn học tốt !!!!

=3(a-b)(b-c)(c-a) nha bn

=\(\frac{1.2.3...30.31}{2\left(2.3.4...31\right).64}=\frac{1}{128}\)

\(Q=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ac+1}\)

    \(=\frac{1.c}{\left(a+ab+1\right)c}+\frac{1.ac}{\left(b+bc+1\right).ac}+\frac{1}{c+ac+1}\)

    \(=\frac{c}{ac+abc+c}+\frac{ac}{abc+abc^2+ac}+\frac{1}{c+ac+1}\)

    \(=\frac{c}{ac+1+c}+\frac{ac}{1+c+ac}+\frac{1}{c+ac+1}\)

    \(=\frac{c+ac+1}{c+ac+1}=1\)

ban k hiểu chỗ nào vậy

Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

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

Ta có:\(\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a\cdot a^2+a\cdot a^2+a\cdot a^2}{a^3+a^3+a^3}\)\(\Rightarrow\frac{3a^3}{3a^3}=1\)

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

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

\(\Leftrightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ac}\)

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

\(\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

<=> a = b = c

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

ab có gạch đầu ko bn?

Nếu ab là ab thì mk giải thế này:

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

\(\Leftrightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}=\frac{10c+a}{c+a}\)

Theo t/c dãy tỉ số=nhau:

\(\frac{10a+b}{a+b}=\frac{10b+c}{b+c}=\frac{10c+a}{c+a}=\frac{\left(10a+b\right)+\left(10b+c\right)+\left(10c+a\right)}{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\)

\(=\frac{\left(10a+a\right)+\left(10b+b\right)+ \left(10c+c\right)}{\left(a+a\right)+\left(b+b\right)+\left(c+c\right)}=\frac{11a+11b+11c}{2a+2b+2c}=\frac{11\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{11}{2}\)

do đó: \(\frac{10a+b}{a+b}=\frac{11}{2}\Rightarrow\left(10a+b\right).2=11.\left(a+b\right)\Rightarrow20a+2b=11a+11b\)

\(\Rightarrow20a-11a=11b-2b\Rightarrow9a=9b\Rightarrow a=b\)

Tương tự với b=c;c=a

=>\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0^3+0^3+0^3=0\)