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$1.若x_1满足2x+2^x=5,x_2满足2x+2\log_{2}{(x-1)} =5,x_1+x_2=(\qquad)$
$解:2^x与\log_{2}{x}是反函数,关于y=x对称.\quad {\color{Red} \because }\quad 2x+2\log_{2}{(x-1)} =5$
$\Rightarrow x+\log_{2}{(x-1)} =\cfrac{5}{2}\Rightarrow \log_{2}{(x-1)} =\cfrac{5}{2}-x{\color{Green} \Rightarrow\begin{cases} y=\log_{2}{(x-1)}\\y=\cfrac{5}{2}-x\end{cases}} $
$同理2x+2^x=5\Rightarrow x+2^{x-1}=\cfrac{5}{2}\Rightarrow 2^{x-1} =\cfrac{5}{2}-x{\color{Green}\Rightarrow \begin{cases} y=2^{x-1}\\y=\cfrac{5}{2}-x\end{cases}}$
${\color{Red}\because\quad} y=2^{x-1}和y=\log_{2}{(x-1) 关于y=x-1对称},x_1,x_2的中心坐标为直线y=x-1与y=\cfrac{5}{2}-x$

$2.已知x\gt1,方程x-(x-1)2^x=0,x-(x-1)\log _2x=0在区间(1,+\infty)的根分别$
$为a,b,以下结论正确的有:$
$A.b-a=2^a-\log _2b\quad B.\cfrac{1}{a}+\cfrac{1}{b}=1\quad C.a+b\lt 4\quad D.a+b\gt 4$
$解:\begin{cases}b=2^a\\a=\log_2 b\end{cases}\Rightarrow b-a=2^a-a=2^a-\log_2b,故A正确。$
$x=(x-1)2^x\Rightarrow \cfrac{x}{x-1}=2^x\Rightarrow 2^a=\cfrac{a}{a-1}{\color{Red} =b} $
$\cfrac{1}{a}+\cfrac{1}{b}=\cfrac{1}{a}+\cfrac{1}{2^a}=\cfrac{1}{a}+\cfrac{a-1}{a}=1$
$a+b=a+\cfrac{a}{a-1}=a-1+\cfrac{a-1+1}{a-1}+1=2+a-1+\cfrac{1}{a-1}\ge 4$

$3.求\cfrac{2\sin 40+\sin 20}{\cos 30}$

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