Giải pt $a) 9^x - 2^ {x+1} = 2^{x+2} -3^{2x-1}$ $b) 8^{\frac{x}{x+2}}=36.3^{2-x}$
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Đáp án:
a.$x=1$
b.$x=4$
Giải thích các bước giải:
a.$9^x-2^{x+1}=2^{x+2}-3^{2x-1}$
$\rightarrow 3^{2x}+3^{2x-1}=2^{x+2}+2^{x+1}$
$\rightarrow 3.3^{2x-1}+3^{2x-1}=2.2^{x+1}+2^{x+1}$
$\rightarrow 4.3^{2x-1}=3.2^{x+1}$
$\rightarrow 2^2.3^{2x-1}=3.2^{x+1}$
$\rightarrow 3^{2x-2}=2^{x-1}$
$\rightarrow 9^{x-1}=2^{x-1}$
$\rightarrow (\dfrac{9}{2})^{x-1}=1$
$\rightarrow x-1=0$
$\rightarrow x=1$
b.$8^{\dfrac{x}{x+2}}=36.3^{2-x}$
$\rightarrow (2^3)^{\dfrac{x}{x+2}}=2^2.3^2.3^{2-x}$
$\rightarrow 2^{\dfrac{3x}{x+2}-2}=3^{4-x}$
$\rightarrow 2^{\dfrac{x-4}{x+2}}=3^{4-x}$
$\rightarrow 2^{\dfrac{x-4}{x+2}}.3^{x-4}=1$
$\rightarrow (2^{\dfrac{1}{x+2}}.3)^{x-4}=1$
$\rightarrow x-4=0$
$\rightarrow x=4$
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