1) log(√7)(1/7)+3^log₃7=log(7^(1/2)7⁻¹+7=-2log₇7+7=-2+7=5.
2) √(25^(1/log⁶5)+49^(1/log₈7))=√(5^2log₅6+7^2log₇8)=
=√(5^log₅6²+7^log₇8²)=√(6²+8²)=√100=10.
3) 10^(1-lg5)=10*10^(-lg5)=10/10^lg5=10/5=2.
4) log(c)(16c²)=log(c)4⁴+log(c)c²=4log(c)2+2=4*(-3)+2=-10.
Используем формулу:cosa*cosb=(cos(a-b)+ <span>cos(a+b))/2.
Имеем: </span>cos3x*cos6x=(cos(3х-6х)+ cos(3х+6х))/2=(cos(3х)+ cos(9х))/2;
cos4x*cos7x=(cos(4х-7х)+ cos(4х+7х))/2=(cos(3х)+ cos(11х))/2;
значит, (cos(3х)+ cos(9х))/2=(cos(3х)+ cos(11х))/2.Отсюда, cos(9х))/2= =cos(11х))/2, cos(9х) =cos(11х),9х=11х, х=0.
Ответ: 0