i dell bt

Sách bài tập lớp 9 ak

Ta có: \(\left(\sqrt{x^2+1}-x\right)\left(\sqrt{y^2+1}-y\right)=\frac{1}{2}\)

=>\(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+xy-x\sqrt{y^2+1}-y\sqrt{x^2+1}=\frac{1}{2}\left(1\right)\)

Lại có: \(\left(\sqrt{x^2+1}-x\right)\left(\sqrt{y^2+1}-y\right)=\frac{1}{2}\)

=>\(\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}.\frac{y^2+1-y^2}{\sqrt{y^2+1}+y}=\frac{1}{2}\)

=>\(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}+y\right)=2\)

=>\(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+xy+x\sqrt{y^2+1}+y\sqrt{x^2+1}=2\left(2\right)\)

Lấy (1)+(2) ta đc:

\(2\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+2xy=\frac{5}{2}\)

=>\(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\frac{5}{4}-xy\)

=>\(x^2y^2+x^2+y^2+1=\frac{25}{16}-\frac{5}{2}xy+x^2y^2\)

=>\(x^2+y^2+\frac{5}{2}xy=\frac{9}{16}\)

=>\(\left(x+y\right)^2+\frac{1}{2}xy=\frac{9}{16}\)

Vì \(\frac{1}{2}xy\le\frac{\left(x+y\right)^2}{8}\)

=>\(\frac{9}{8}.\left(x+y\right)^2\ge\frac{9}{16}\)

=>\(x+y\ge\frac{1}{\sqrt{2}}\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{\sqrt{2}}\)

Vậy Min \(F=\frac{1}{\sqrt{2}}< =>x=y=\frac{1}{2\sqrt{2}}\)

\(M=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-4}}{y}\)

ta co \(1.\sqrt{x-1}\le\frac{x+1-1}{2}=\frac{x}{2}\)

\(2.\sqrt{y-4}=\sqrt{4}\sqrt{y-4}\le\frac{y-4+4}{2}=\frac{y}{2}\)

\(M=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{4}\sqrt{y-4}}{2y}\le\frac{\frac{x}{2}}{x}+\frac{\frac{y}{2}}{2y}=\frac{x}{2x}+\frac{y}{4y}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

vay max \(M=\frac{3}{4}\)khi \(\hept{\begin{cases}x=2\\y=8\end{cases}}\)

x=2 y=8