Question

Заранее . решить уравнения 1) 2) 3) 4)

Answer 1

1) $5^{x+3} -5^{x+2}=3^{x+4}-7*3^{x+1}$
$5^{x+1}(25-5 )=3^{x+1}(27-7)$
$( \frac{5}{3})^{x+1} = \frac{20}{20} =1$
$x+1=0$
$x=-1$

2)$( \sqrt{2+ \sqrt{3} } ) ^{x} =t, t\ \textgreater \ 0$
$( \sqrt{2- \sqrt{3} } ) ^{x}= \frac{1}{t}$
$\frac{1}{t} +t=4$
$t^{2} -4t+1=0$
$t_{1} = 2+\sqrt{4-1}=2+ \sqrt{3}$
$(2+ \sqrt{3}) ^{ \frac{x}{2} } = (2+ \sqrt{3})^1 }$
$\frac{x}{2} =1$
$x=2$
$t_{2} =2- \sqrt{3}$
$(2- \sqrt{3}) ^{ \frac{x}{2} } = (2- \sqrt{3})^{-1}$
$\frac{x}{2} =-1$
$x=-2$
$x_{1}=-2 , x_{2} =2$

3)$3* 2^{2x}-5 (2*3)^{x} +2* 3^{2x}=0 |/ 3^{2x}$
$3* ( \frac{2}{3} )^{2x} -5*( \frac{2}{3} )^{x}+2=0$
$( \frac{2}{3} )^{x}=t, t\ \textgreater \ 0$
$( \frac{2}{3} )^{2x}= t^{2}$
$3t-5t+2=0$
$t_{1} = \frac{2}{3}$
$( \frac{2}{3} )^{x} = \frac{2}{3}$
$x=1$
$t_{2} =1$
$( \frac{2}{3} )^{x} =(\frac{2}{3}) ^{0}$
$x=0$
$x_{1}=0 , x_{2} =1$

4)$125* 5^{2x}-70*(2*5)^{x} +8* 2^{2x}=0 |/ 5^{2x}$
$125-70*( \frac{2}{5} )^{x}+8* ( \frac{2}{5} )^{2x} =0$
$( \frac{2}{5} )^{x}=t, t\ \textgreater \ 0$
$( \frac{2}{5} )^{2x}= t^{2}$
$8 t^{2}-70t+125=0$
$t_{1} = \frac{5}{2}$
$( \frac{2}{5} )^{x} = (\frac{2}{5}) ^{-1}$
$x=-1$
$t_{2} = \frac{25}{4}$
$( \frac{2}{5} )^{x} = (\frac{2}{5}) ^{-2}$
$x=-2$
$x_{1}=-2 , x_{2} =-1$