Question

Помогите!!! Срочно. 7*9^(x^2-3x+1) + 5*6^(x^2-3x+1) - 48*4^(x^2-3x)

Answer 1

$7*9^{x^2-3x+1} + 5*6^{x^2-3x+1} - 48*4^{x^2-3x}=0\7*3^2*3^{2(x^2-3x)}+5*6*2^{x^2-3x}*3^{x^2-3x}-48*2^{2(x^2-3x)}=0\63*3^{2(x^2-3x)}+30*2^{x^2-3x}*3^{x^2-3x}-48*2^{2(x^2-3x)}=0|:3\21*3^{2(x^2-3x)}+10*2^{x^2-3x}*3^{x^2-3x}-16*2^{2(x^2-3x)}=0|:2^{2(x^2-3x)}\21*(frac{3}{2})^{2(x^2-3x)}+10*(frac{3}{2})^{x^2-3x}-16=0\ (frac{3}{2})^{x^2-3x}=a$

$21a^2+10a-16=0\D:(frac{10}{2})^2+21*16=361;\a_1,_2=frac{-5pm19}{21}, qquad a_1=frac{2}{3}, quad a_2=-frac{8}{7};\\(frac{3}{2})^{x^2-3x}=frac{2}{3}\(frac{3}{2})^{x^2-3x}=(frac{3}{2})^{-1}\x^2-3x=-1\x^2-3x+1=0\D:9-4=5\x_1,_2=frac{3pmsqrt{5}}{2};\\(frac{3}{2})^{x^2-3x} neq -frac{8}{7}.$