36-(-(9c-15))=3(3c+7)
36+9c-15=9c+21
9c-9c=21-36+15
0=0
У/4=у-1
у=4(у-1)
у=4у-4
у-4у=-4
-3у=-4
у=4/3=1 1/3
Решение
2lg5 + 1/2lg16 = lg(5^2) + lg√(16) = lg(25*4) = lg100 = lg(10)^2 = 2lg10 = 2
3^x*(27+1)=2^x*(5*16-17)
3^x*28=2^x*63
(3/2)^x=63/28=9/4
x=2
1)2d/(√3-a)=2d(√3+a)/(√3-a)(√3+a)=(2d√3+2ad)/(3-a^2)
2)3x/(√x+2)=3x(√x-2)/(√x+2)(√x-2)=(3x√x-6x)/(x-4)
3)5t/(√t-√s)=5t(√t+√s)/(√t+√s)(√t-√s)=(5t√t+5t√s)/(t-s)
4)7m/(√m-√(2n)=7m(√m+√(2n))/(√m-√(2n))(√m+√(2n))=(7m√m+7m√(2n))/(m-2n)