Giải thích các bước giải:
1.Ta có :
$2^{x+1}.4^{x-1}.\dfrac{1}{8^{1-x}}=16^x$
$\to 2^{x+1}.2^{2(x-1)}.\dfrac{1}{2^{3(1-x)}}=2^{4x}$
$\to 2^{x+1}.2^{2(x-1)}.2^{-3(1-x)}=2^{4x}$
$\to 2^{x+1+2(x-1)-3(1-x)}=2^{4x}$
$\to 2^{6x-4}=2^{4x}$
$\to 6x-4=4x$
$\to x=2$
2.Ta có :
$(\sqrt{10}+3)^{\dfrac{x-3}{x-1}}=(\sqrt{10}-3)^{\dfrac{x+1}{x+3}}$
$\to (\sqrt{10}+3)^{\dfrac{x-3}{x-1}}.(\sqrt{10}-3)^{\dfrac{x-3}{x-1}}=(\sqrt{10}-3)^{\dfrac{x+1}{x+3}}.(\sqrt{10}-3)^{\dfrac{x-3}{x-1}}$
$\to ((\sqrt{10}+3)(\sqrt{10}-3))^{\dfrac{x-3}{x-1}}=(\sqrt{10}-3)^{\dfrac{x+1}{x+3}+\dfrac{x-3}{x-1}}$
$\to (10-9)^{\dfrac{x-3}{x-1}}=(\sqrt{10}-3)^{\dfrac{x+1}{x+3}+\dfrac{x-3}{x-1}}$
$\to 1^{\dfrac{x-3}{x-1}}=(\sqrt{10}-3)^{\dfrac{x+1}{x+3}+\dfrac{x-3}{x-1}}$
$\to \dfrac{x+1}{x+3}+\dfrac{x-3}{x-1}=0$
$\to \left(x+1\right)\left(x-1\right)+\left(x-3\right)\left(x+3\right)=0$
$\to 2x^2-10=0$
$\to x=\pm\sqrt 5$
7.$(\dfrac 13)^{\dfrac 2x}+3(\dfrac 13)^{\dfrac 1x+1}=12$
$\to ((\dfrac 13)^{\dfrac 1x})^2+3.\dfrac 13^{\dfrac 1x}.\dfrac 13=12$
$\to ((\dfrac 13)^{\dfrac 1x})^2+\dfrac 13^{\dfrac 1x}-12=0$
$\to ((\dfrac 13)^{\dfrac 1x}-3)((\dfrac 13)^{\dfrac 1x}+4)=0$
$\to (\dfrac 13)^{\dfrac 1x}-3=0$
$\to (\dfrac 13)^{\dfrac 1x}=3$
$\to \dfrac 1x=-1$
$\to x=-1$
