Đáp án: $ 2<m<4$
Giải thích các bước giải:
Ta có:
$6^x+(3-m)2^x-m=0$
$\to 6^x+3\cdot 2^x-m\cdot 2^x-m=0$
$\to 6^x+3\cdot 2^x-m(\cdot 2^x+1)=0$
$\to 6^x+3\cdot 2^x=m(\cdot 2^x+1)$
$\to m=\dfrac{6^x+3\cdot 2^x}{2^x+1}$
Đặt
$f(x)=\dfrac{6^x+3\cdot 2^x}{2^x+1}$
$\to f(x)-2=\dfrac{6^x+3\cdot 2^x}{2^x+1}-2$
$\to f(x)-2=\dfrac{6^x+3\cdot 2^x-2(2^x+1)}{2^x+1}$
$\to f(x)-2=\dfrac{6^x+3\cdot 2^x-2\cdot 2^x-2}{2^x+1}$
$\to f(x)-2=\dfrac{6^x+\cdot 2^x-2}{2^x+1}$
Vì $x>0\to 6^x+2^x-2>6^0+2^0-2=0$
$\to \dfrac{6^x+\cdot 2^x-2}{2^x+1}>0$ vì $2^x+1>0$
$\to f(x)-2>0$
$\to f(x)>2(1)$
Ta có:
$f(x)-4=\dfrac{6^x+3\cdot 2^x}{2^x+1}-4$
$\to f(x)-4=\dfrac{6^x+3\cdot 2^x-4(2^x+1)}{2^x+1}$
$\to f(x)-4=\dfrac{6^x+3\cdot 2^x-4\cdot 2^x-4}{2^x+1}$
$\to f(x)-4=\dfrac{6^x- 2^x-4}{2^x+1}$
Do $0<x<1\to 6^x=(2\cdot 3)^x=2^x\cdot 3^x<2^x\cdot 3^1=3\cdot 2^x$
$\to 6^x- 2^x-4<3\cdot 2^x-2^x-4=2\cdot 2^x-4<2\cdot 2^1-4=0$
$\to \dfrac{6^x- 2^x-4}{2^x+1}<0$ vì $2^x+1>0$
$\to f(x)-4<0$
$\to f(x)<4(2)$
Từ $(1), (2)\to 2<f(x)<4$
$\to$Để phương trình có nghiệm $\to 2<m<4$
