(1)⇔log2(x2+1)−(log22+log2(y2+1))=−1⇔log2(x2+1)−log2(y2+1)=0⇔log2x2+1y2+1=0⇔x2+1=y2+1⇔x=yVới x=y(2)⇔43x+5.23x−6=0Đặt t=23x>0(2)⇒t2+5t−6=0⇔t=1⇔23x=1⇔x=0Vậy nghiệm hệ (x;y)=(0;0)
(1)⇔log2(x2+1)−(log22+log2(y2+1))=−1⇔log2(x2+1)−log2(y2+1)=0$\Leftrightarrow \log_2\frac{x^2+1}{y^2+1}=0\Leftrightarrow x^2+1=y^2+1\Leftrightarrow x=
\pm y
Vớix=y
(2)⇔43x+5.23x−6=0$Đặt$t=23x>0(2)\Rightarrow t^2+5t-6=0\Leftrightarrow t=1\Leftrightarrow 2^{3x}=1\Leftrightarrow x=0$
Với x=−yTương tựVậy nghiệm hệ
(x;y)=(0;0)