a+b=2=> a=2-b
\(\Rightarrow\left(1-\frac{4}{a^2}\right)\left(1-\frac{4}{b^2}\right)=\left(\frac{a^2-4}{a^2}\right)\left(\frac{b^2-4}{b^2}\right)=\frac{\left(2-b\right)^2-4}{\left(2-b\right)^2}.\frac{b^2-4}{b^2}\)
=\(\frac{b^2-2b-8}{b^2-2b}\)
đặt A=\(\frac{b^2-2b-8}{b^2-2b}\)
đkxđ \(\hept{\begin{cases}b\ne0\\b\ne2\end{cases}}\)
\(\Leftrightarrow Ab^2-2bA=b^2-2b-8\)
\(\Leftrightarrow\left(A-1\right)b^2-2\left(A-1\right)b+8=0\)
nếu A=1 => 8=0 (vô lý)
nếu A khác 1 pt có nghiệm khi \(\Delta\ge0\Leftrightarrow\left[-2\left(A-1\right)\right]^2-4\left(A-1\right).8\ge0\)
\(4A^2-40A+36\ge0\Leftrightarrow A^2-10A+9\ge0\Leftrightarrow\hept{\begin{cases}A\le1\\A\ge9\end{cases}}\)
GTNN A=9 dấu "=" <=> a=b=1
bạn ơi mình đặt nhầm B thành A rồi bn tự sửa lại nhé!
\(B=\left(1-\frac{4}{a^2}\right)\left(1-\frac{4}{b^2}\right)=\left(1-\frac{2}{a}\right)\left(1-\frac{2}{b}\right)\left(1+\frac{2}{a}\right)\left(1+\frac{2}{b}\right)\)
\(=\frac{\left(2-a\right)\left(2-b\right)\left(a+2\right)\left(b+2\right)}{a^2b^2}=\frac{ab.\left(a+2\right)\left(b+2\right)}{a^2b^2}=\frac{ab+2\left(a+b\right)+4}{ab}=\frac{8}{ab}+1\)
Theo BĐT Cauchy thì : \(a+b\ge2\sqrt{ab}\Rightarrow ab\le\frac{\left(a+b\right)^2}{4}\)
Suy ra : \(A\ge\frac{8}{\frac{2^2}{4}}+1=9\).Đẳng thức xảy ra khi a = b = 1/2
Vậy ......................................
Đẳng thức xảy ra khi a = b = 1 nhé.
