Question

помогите пожалуйста....Очень срочно! пожалуйста
5^(2x+5)*9^(2-x)*64^(x+1)=600^(4x+1)

Answer 1

$5^{(2x+5)}*9^{(2-x)}*64^{(x+1)}=600^{(4x+1)} \ \ 5^{2x}*5^5*frac{9^2}{9^x}*64^x*64^1=600^{4x}*600^1 \ \ 5^{2x}*3125*frac{81}{3^{2x}}*8^{2x}*64=360000^{2x}*600 \ \ 5^{2x}*3125*81*(frac{1}{3})^{2x}*8^{2x}*64=360000^{2x}*600 \ \ 5^{2x}*(frac{1}{3})^{2x}*8^{2x}*(3125*81*64)=360000^{2x}*600 \ \ 5^{2x}*(frac{1}{3})^{2x}*8^{2x}*16200000=360000^{2x}*600$

$5^{2x}*(frac{1}{3})^{2x}*8^{2x}*27000=360000^{2x} \ \ (5*(frac{1}{3})*8)^{2x}*27000=360000^{2x} \ \ (frac{40}{3})^{2x}*27000=360000^{2x} \ \ 27000= (frac{360000}{frac{40}{3}})^{2x} \ \ 27000= (frac{360000*3}{40})^{2x} \ \ 27000= 27000^{2x} \ \ 2x=1 \ \ x= frac{1}{2}$