Question

1. Cho \(a,b,c\ne0\) và \(a+b+c=0\) . Tính gt of bt:

\(Q=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)

2. Cho hai số thực a, b thoả mãn a>b và ab=2. Tìm GTNN của bt \(M=\frac{a^2+b^2}{a-b}\)

Answer 1

\(\frac{1}{a^2+b^2-c^2}=\frac{1}{a^2+\left(b-c\right)\left(b+c\right)}=\frac{1}{a^2-a\left(b-c\right)}=\frac{1}{a\left(a-b+c\right)}=\frac{1}{-2ab}\)

Tương tự \(\Rightarrow Q=-\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0\)

\(M=\frac{\left(a-b\right)^2+2ab}{a-b}=a-b+\frac{4}{a-b}\ge2\sqrt{\frac{4\left(a-b\right)}{a-b}}=4\)

\(M_{min}=4\) khi \(\left\{{}\begin{matrix}a-b=2\\ab=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{3}+1\\b=\sqrt{3}-1\end{matrix}\right.\)