Question

log0,4x+log0,4(x+1)=log0,4(8-x)

Answer 1

Ответ: 2

Объяснение:

$log_{0,4}x+log_{0,4}(x+1)=log_{0,4}(8-x)$

ОДЗ: $\left \{ {{x > 0} \atop {8-x>0}} \right.\left \{ {{x>0} \atop {x<8}} \right. \Rightarrow x\in(0; 8)$

$log_{0,4}(x(x+1))=log_{0,4}(8-x)\\x(x+1)=8-x\\x^2+x-8+x=0\\x^2+2x-8=0\\\left \{ {{x_1+x_2=-2} \atop {x_1x_2=-8}} \right. \Rightarrow x_1=2, x_2=-4\notin OD3 \\x=2$