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Question

$\begin{cases}2^x+x=y+log_2y \\ log_2x+y=5 \end{cases}$

khangnguyenthanh · Answer 1 · 5/31/2014 5:40:40 PM

Xét hàm:

$f(t)=2^t+t$ .

Ta có:

$f'(t)=2^t\ln2+1>0,\forall t>0$

Suy ra

$f(t)$ đồng biến trên khoảng

$(0;+\infty)$ .

Xét hàm:

$g(t)=2^t+\log_2t$ .

Ta có:

$g'(t)=2^t\ln2+\dfrac{1}{t\ln2}>0,\forall t>0$

Suy ra

$g(t)$ đồng biến trên khoảng

$(0;+\infty)$ .

Ta có:

$\left\{\begin{array}{l}2^x+x=y+\log_2y\\\log_2x+y=5\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}f(x)=f(\log_2y)\\\log_2x+y=5\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}x=\log_2y\\\log_2x+y=5\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}y=2^x\\\log_2x+2^x=5\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}y=2^x\\g(x)=g(2)\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}x=2\\y=4\end{array}\right.$

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