câu hỏi ko tl cx thấy xàm xàm xàm xmà

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.

Ta có

\(M=\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)=2+\frac{a}{b}+\frac{b}{a}+a+b+\frac{1}{a}+\frac{1}{b}\)

\(\ge2+2+a+b+\frac{4}{a+b}\)

\(=4+a+b+\frac{2}{a+b}+\frac{2}{a+b}\)

 \(\ge4+2\sqrt{\left(a+b\right).\frac{2}{\left(a+b\right)}}+\frac{2}{\sqrt{2\left(a^2+b^2\right)}}\)

\(=4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)

M=\(\frac{a^4}{a\left(b+1\right)^2}+\frac{b^4}{b\left(a+1\right)^2}\)

áp dụng bdt bunhiacopxki ta co

(a+b)M>=\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2\)

\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2>=\left[\frac{\left(a+b^2\right)}{a+1+b+1}\right]^2\)

\(=\frac{\left(a+b\right)^4}{\left(a+b+2\right)^2}>=\frac{\left(a+b\right)^4}{4\left(a+b\right)^2}\)(do 2<=a+b)

=\(\frac{\left(a+b\right)^2}{4}\)

do do M(a+b)>=\(\frac{\left(a+b\right)^2}{4}\)

=>M>=\(\frac{a+b}{4}>=\frac{1}{2}\)

dau = xay ra <=> a=b=1

\(A=\left(a+b+1\right)\left(a^2+b^2\right)+\frac{4}{a+b}+1-1\ge\left(a+b+1\right)2\sqrt{\left(ab\right)^2}+\frac{\left(2+1\right)^2}{a+b+1}-1\)

\(=2\left(a+b+1\right)+\frac{9}{a+b+1}-1\ge2\sqrt{ab}+1+2\sqrt{\frac{9\left(a+b+1\right)}{a+b+1}}-1\ge2+6=8\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}a^2=b^2\left(1\right)\\\frac{2}{a+b}=1\left(2\right)\\a+b+1=\frac{9}{a+b+1}\left(3\right)\end{cases}}\)

pt \(\left(1\right)\)\(\Leftrightarrow\)\(a=b\) ( vì a, b > 0 ) 

pt \(\left(2\right)\)\(\Leftrightarrow\)\(a=b=1\)

pt \(\left(3\right)\)\(\Leftrightarrow\)\(\left(a+b+1\right)^2=9\)\(\Leftrightarrow\)\(a+b+1=3\) ( đúng vì \(a=b=1\) ) 

Vậy GTNN của \(A\) là \(8\) khi \(a=b=1\)

Chúc bạn học tốt ~