Question

Cho ab+a+b=1.Chứng minh (a^2 + 1)(b^2 + 1)=2(a+b)^2

Answer 1

Ta có : \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+ab+a+b\right)\left(b^2+ab+a+b\right)\)

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

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

Answer 2

chịu ai bt đc 90% là 2k10 mà

Answer 3

Ta có:\(\left(a^2+1\right)\left(b^2+1\right)=2\left(a+b\right)^2\)

Xét \(VT=\left(a^2+1\right)\left(b^2+1\right)\)

\(=\left(a^2+ab+a+b\right)\left(b^2+ab+a+b\right)\)Do \(ab+a+b=1\)

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

\(=\left(a+b\right)\left(a+1\right)\left(b+1\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\left(ab+a+b+1\right)\)Do \(ab+a+b=1\)

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

\(=2\left(a+b\right)^2=VP\)