log9(x+2)2+log3√x+8−−−−√=2+log27(4−x)3
Đk tự làm nhé
$\Leftrightarrow \dfrac{1}{2}\log_3 (x+2)^2 + 2\log_3 \sqrt{x+8} = \log_3 9 +\dfrac{1}{3}\log_3 (4-x)^3$
$\Leftrightarrow \log_3 (x+2) + \log_3 (x+8) = \log_3 9 +\log_3 (4-x)$
$\Leftrightarrow \log_3 (x+2)(x+8) = \log_3 9(4-x)$
$\Leftrightarrow (x+2)(x+8)=9(4-x)$ giải ra được $x = 1;\ x = -20(LOAI)$