Question

11 класс. Уравнения. Задание на картинке. 40 баллов.

Answer 1

Не пишу исходные.

$( 3^{x-3} ) ^{x+4} = 3^{x-3} log(3)(x - 3)(x + 4) = log(3)(3 - x) log(3)x^2 + log(3)x - 12log(3) = log(3)(3 - x) log(3)x^2 + log(3)x - 12log(3) = 3log(3) - log(3)x log(3)x^2 + 2log(3)x - 15 log(3) = 0 log(3)(x - 3)(x + 5) = 0 x - 3 = 0 ; x + 5 = 0 x1 = 3 ; x2 = - 5$

$frac{7}{4} = 7 * 2^{x^2-x-4}$
$2^{x^2-x-4} = frac{1}{4} x^2 - x - 4 = - 2 x^2 - x = 2$
$x^2 - x + frac{1}{4} = frac{9}{4}$
$(x - frac{1}{2} )^2 = frac{9}{4}$
$x - frac{1}{2} = 1,5 ; x - frac{1}{2} = - 1,5 x1 = 2 ; x2 = -1$

Answer 2

$(3^{x-3})^{x+4}=( frac{1}{3})^{3x-1} \ 3^{x^2+x-12}=3^{-3x+1} \ x^2+x-12=-3x+1 \ x^2+4x-13=0 \ Delta=16+4*13=16+52=68=4*17 \ x_1= frac{-4-2 sqrt{17} }{2} =-2- sqrt{17} \ x_2=-2+ sqrt{17}$
$2^{x+2}+2^{x+3}+2^{x+4}=7*2^{x^2} \ 2^x(4+8+16)=7*2^{x^2} \ 28*2^x=7*2^{x^2} \ 4*2^x=2^{x^2} \ 2^{x+2}=2^{x^2} \ x^2=x+2 \ x^2-x-2=0 \ Delta=1+8=9=3^2 \ x_1= frac{1-3}{2}=-1 \ x_2= frac{1+3}{2}=2$