ai có cách hay thì giải cho tui xem vs na
pt: [TEX]4^x-2^{x+1}+2(2^x-1)sin(2^x+y-1)+2=0[/TEX]
\Leftrightarrow[TEX]({2}^{2x} -2.2^x +1) +2(2^x-1)sin(2^x+y-1) +sin^2(2^x+y-1) +cos^2(2^x+y-1) =0[/TEX]
\Leftrightarrow[TEX](2^x -1 +sin(2^x-1+y))^2 +cos^2(2^x-1+y) =0[/TEX]
\Leftrightarrow[TEX]2^x-1+sin(2^x-1+y)=0 [/TEX]và[TEX]cos^2(2^x-1+y)=0[/TEX]
\Leftrightarrow[TEX]2^x -1+sin(2^x-1+y) =0[/TEX] và[TEX] sin(2^x-1+y) =\pm1[/TEX]
Th1:[TEX]2^x -1+sin(2^x-1+y) =0[/TEX] và[TEX] sin(2^x-1+y) =1[/TEX]\Rightarrow[TEX]2^x =0[/TEX]\Rightarrowvô nghiệm
Th2:[TEX]2^x -1+sin(2^x-1+y) =0[/TEX] và[TEX] sin(2^x-1+y) =-1[/TEX]\Leftrightarrow[TEX]2^x =2[/TEX]và [TEX]sin(1+y) =-1[/TEX]\Leftrightarrow[TEX]x =1[/TEX]và[TEX]y =-\frac{\pi}{2} -1 +k2\pi[/TEX]