Question

[Nguyệt Hà]

Cho a,b là số hữu tỉ  tm \(\left(a^2+b^2-2\right)\left(a+b\right)^2+\left(1-ab\right)^2=-4ab\)

CM\(\sqrt{1+ab}\) là số hữu tỉ

Answer 1

[Vũ Tiến Manh]

<=> (a²+b²)(a+b)²- 2(a+b)² +1+ a²b² -2ab= -4ab <=> (a²+b²)(a²+b²+2ab)- 2(a+b)²+ a²b²+ 2ab+ 1=0

<=> [(a²+b²)²+(a²+b²).2ab+a²b² ] - 2(a²+b²+2ab)+2ab+1=0 <=> (a²+b²+ab)²- 2(a²+b²+ab)+1=0

<=> (a²+b²+ab-1)²=0 <=> a²+b²+ab-1=0 <=> (a+b)²-(1+ab)=0 <=> (a+b)² =1+ab => \(\sqrt{1+ab}=\)\(|a+b|\)là số hữu tỉ

Answer 2

[Tran Le Khanh Linh]

\(\left(GT\right)\Rightarrow\left[\left(a+b\right)^2-2\left(ab+1\right)\right]\left(a+b\right)^2+\left(1+ab\right)^2=0\)

\(\Leftrightarrow\left(a+b\right)^4-2\left(a+b\right)^2\left(1+ab\right)+\left(1+ab\right)^2=0\)

\(\Leftrightarrow\left[\left(a+b\right)^2-\left(1+ab\right)\right]^2=0\Rightarrow\left(a+b\right)^2-\left(1+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)^2=1+ab\Leftrightarrow\left|a+b\right|=\sqrt{1+ab}\left(a,b\inℚ\right)\)