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乌贼
发表于 2013-12-27 13:25
$5^x+2^y=\dfrac{7}{10}\riff5^x=\dfrac{7}{10}-2^y$
令$f_1(x)=5^x,g_1(y)=\dfrac{7}{10}-2^y$,知方程$5^x+2^y=\dfrac{7}{10}$的解即为$f_1,g_2$的交点(有唯一解)
同理
$5^y+2^x=\dfrac{7}{10}\riff5^y=\dfrac{7}{10}-2^x$
令$f_2(y)=5^y,g_2(y)=\dfrac{7}{10}-2^x$,知方程$5^y+2^x=\dfrac{7}{10}$的解即为$f_2,g_2$的交点(有唯一解)
$f_1=f_2,g_1=g_2$有$x=y$…… |
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