Question

Chứng minh rằng nếu a,b,c,d thỏa mãn đẳng thức:

[ab(ab-2cd)+c\(^2\)d\(^2\)][ab(ab-2)+2(ab+1)] bằng 0 thì chúng lập thành một tỉ lệ thức.

Answer 1

\(\left[ab.\left(ab-2cd\right)+c^2d^2\right].\left[ab.\left(ab-2\right)+2.\left(ab+1\right)\right]=0\)

\(\Rightarrow\left(a^2b^2-2abcd+c^2d^2\right).\left(a^2b^2-2ab+2ab+2\right)=0\)

\(\Rightarrow\left(a^2b^2-abcd-abcd+c^2d^2\right).\left(a^2b^2+2\right)=0\)

\(\Rightarrow\left[\left(a^2b^2-abcd\right)-\left(abcd-c^2d^2\right)\right].\left(a^2b^2+2\right)=0\)

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

\(\Rightarrow\left(ab-cd\right).\left(ab-cd\right).\left(a^2b^2+2\right)=0\)

\(\Rightarrow\left(ab-cd\right)^2.\left(a^2b^2+2\right)=0\)

Vì \(a^2b^2+2>0\) \(\forall x.\)

\(\Rightarrow\left(ab-cd\right)^2=0\)

\(\Rightarrow ab-cd=0\)

\(\Rightarrow ab=0+cd\)

\(\Rightarrow ab=cd.\)

\(\Rightarrow\frac{a}{c}=\frac{d}{b}\left(đpcm\right).\)

Chúc bạn học tốt!