Đặt \(a-1=x>0,b-1=y>0\), ta có

\(A=\frac{\left(x+1\right)^2}{x}+\frac{\left(y+1^2\right)}{y}=\frac{x^2+2x+1}{x}+\frac{y^2+2y+1}{y}\)

\(=\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+4\)

Với \(x>0,y>0\)ta có \(x+\frac{1}{x}\ge2,y+\frac{1}{y}\ge2\)nên \(A\ge8\)

\(Min_A=8\Leftrightarrow x=y=1\Leftrightarrow a=b=2\)

P/s tham khảo nha

Sử dụng \(AM-GM\)ta có :

\(\frac{a^2}{a-1}+4\left(a-1\right)\ge2\sqrt{\left(2a\right)^2}=4a\)

Tương tự : \(\frac{b^2}{b-1}+4\left(b-1\right)\ge4b\)

Cộng theo vế : \(A+4\left(a+b\right)-8\ge4\left(a+b\right)\)

\(< =>A\ge8\)

Dấu = xảy ra \(< =>a=b=2\)

Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có :

\(\left[\left(\sqrt{\frac{2}{1-x}}\right)^2+\left(\sqrt{\frac{1}{x}}\right)^2\right]\left[\sqrt{1-x}^2+\sqrt{x}^2\right]\ge\left(\sqrt{\frac{2}{1-x}}.\sqrt{1-x}+\sqrt{\frac{1}{x}}.\sqrt{x}\right)^2\)

\(\Rightarrow\left(\frac{2}{1-x}+\frac{1}{x}\right)\left(1-x+x\right)\ge\left(\sqrt{2}+\sqrt{1}\right)^2\Rightarrow A\ge3+2\sqrt{2}\)

Dấu "=" xảy ra khi \(x=\sqrt{2}-1\)

Ta phải có \(\left|x\right|\le\sqrt{3}\).Dễ thấy \(A>0\).Ta xét biểu thức

\(B=\frac{1}{A}=2-\sqrt{3-x^2}\)

Ta có:

\(0\le\sqrt{3-x^2}\le\sqrt{3}\Rightarrow-\sqrt{3}\le-\sqrt{3-x^2}\le0\)

\(\Rightarrow2-\sqrt{3}\le2-\sqrt{3-x^2}\le2\)

\(Min_B=2-\sqrt{3}\Leftrightarrow\sqrt{3}=\sqrt{3-x^2}\Leftrightarrow x=0\)

Khi đó \(Max_A=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\)

\(Max_B=2\Leftrightarrow\sqrt{3-x^2}=0\Leftrightarrow x=\pm\sqrt{3}\)

Khi đó \(Min_A=\frac{1}{2}\)

P/s tham khảo nha

ok bạn

mk nha k cho mk vs mk hok lp 6