Question

Cho a,b,c là 3 số nguyên khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)

Timhs gt bt \(A=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)

Answer 1

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

\(A=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Answer 2

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow1=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).abc\Leftrightarrow1=bc+ac+ab\)

\(A=\left(bc+ac+ab+a^2\right)\left(bc+ac+ab+b^2\right)\left(bc+ac+ab+c^2\right)\)

\(A=\left[c\left(a+b\right)+a\left(a+b\right)\right]\left[c\left(a+b\right)+b\left(a+b\right)\right]\left[c\left(c+b\right)+a\left(c+b\right)\right]\)

\(A=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(A=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)