Question

Giải pt :

a, \(32^{\dfrac{x+5}{x-7}}=0,25\cdot128^{\dfrac{x-7}{x+3}}\)

b, \(\left(3+2x-x^2\right)^{\sqrt{3}\sin x}=\left(3+2x-x^2\right)^{1-cosx}\)

Answer 1

a, \(32^{\dfrac{x+5}{x-7}}=0,25\cdot128^{\dfrac{x-7}{x+3}}\)

\(\Leftrightarrow2^{5\cdot\dfrac{x+5}{x-7}}=2^{-2}\cdot2^{7\cdot\left(\dfrac{x-7}{x+2}\right)}\)

\(\Leftrightarrow\dfrac{5x+25}{x-7}=-2+7\cdot\left(\dfrac{x-7}{x+3}\right)\)

\(\Leftrightarrow\dfrac{5x25}{x-7}=-2\cdot\dfrac{7x-49}{x+3}\)

\(\Leftrightarrow\dfrac{5x+25}{x-7}=\dfrac{-2x-6+7x-49}{x+3}\)

\(\Leftrightarrow\dfrac{5x+25}{x-7}=\dfrac{5x-55}{x-3}\)

\(\Rightarrow5x^2+15x+25x+75=5x^2-35x-55x+385\)

\(\Leftrightarrow130x+310=0\)

\(\Leftrightarrow x=\dfrac{31}{13}\)