Question

Answer 1

Chọn B

Answer 2

\(\Leftrightarrow4^{1+x}+4^{1-x}=2\left(m+1\right)\left(2^{1+x}-2^{1-x}\right)+16-8x\)

Đặt \(2^{1+x}-2^{1-x}=t\in\left[0;3\right]\Rightarrow t^2=4^{1+x}+4^{1-x}-8\)

Phương trình trở thành:

\(t^2+8=2\left(m+1\right)t+16-8m\)

\(\Rightarrow t^2-2t-8=2m\left(t-4\right)\)

\(\Leftrightarrow\left(t+2\right)\left(t-4\right)=m\left(t-4\right)\)

\(\Rightarrow t+2=m\)

\(\Rightarrow2\le m\le5\)