$计算:\sqrt{1.777777777\cdots}$
$法一:令原式为x,x^2=1.777777777\cdots$
$10x^2=17.77777777\cdots\Rightarrow 9x^2=16,x=\cfrac{4}{3}$
$法二:观察\cfrac{1}{9}=0.1111111\cdots$
$\Rightarrow 0.7777777\cdots=\cfrac{7}{9},1.777777777\cdots=1+\cfrac{7}{9},$
$\therefore \sqrt{1.777777777\cdots}\sqrt{\cfrac{16}{9}}=\cfrac{4}{3}$

$已知不等式2x-m\ln x+1\ge 2\ln x+n(m,n\in \mathbb{R} ),且m\ne -2),对于任意正实数x恒成立$
$则\cfrac{n-5}{m+2}的最大值为(\qquad)$
$对数单身狗,2x-m\ln x+1\ge 2\ln x+n\Rightarrow 2x-n+1\ge 2\ln x+m\ln x$
$(m+2)\ln x\le 2x+1-n\quad \because \quad 2x+1-n是单调递函数\quad\therefore \quad m+2\gt 0$
$\Rightarrow \ln x\le \cfrac{2x+1-n}{m+2},观察右边式子,x=2时,有\cfrac{5-n}{m+2}$
$显然\ln x在直线下方,且在点(2,\cfrac{5-n}{m+2})与直线相切,故\ln 2\le \cfrac{5-n}{m+2}\Rightarrow \cfrac{n-5}{m+2}\le -\ln 2$

$f(x)=x+\sqrt{1-x^2},或f(x)=2x+\sqrt{1-4x^2}的最大值为(\qquad ),或g(x)=x-\sqrt{1-x^2}最小值$