X/(2+3*x)+5/(2-3*x)=5*(2+3*x)/((2-3*x)*(2+3*x)),
x/(2+3*x)+5/(2-3*x)=5/(2-3*x),
x/(2+3*x)=0,
x=0
Ответ: x=0.
Y`=(3sin6x)`*(6x)`-(sin18x)`*(18x)`=3cos6x*6 -cos18x*18=
=18cos6x-18cos18x=18(cos6x-cos18x)
или
18*(-2sin(-6x)sin12x)=36sin6x*sin12x
![0,5^{x^{2}-5x }\leq16\\\\(\frac{1}{2})^{x^{2}-5x }\leq(\frac{1}{2})^{-4}\\\\x^{2} -5x\geq-4\\\\x^{2} -5x+4\geq0\\\\(x-4)(x-1)\geq 0](https://tex.z-dn.net/?f=0%2C5%5E%7Bx%5E%7B2%7D-5x+%7D%5Cleq16%5C%5C%5C%5C%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7Bx%5E%7B2%7D-5x+%7D%5Cleq%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B-4%7D%5C%5C%5C%5Cx%5E%7B2%7D+-5x%5Cgeq-4%5C%5C%5C%5Cx%5E%7B2%7D+-5x%2B4%5Cgeq0%5C%5C%5C%5C%28x-4%29%28x-1%29%5Cgeq+0)
+ - +
_________[1]__________[4]___________
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Ответ : x ∈ (- ∞ , 1] ∪ [4 ; + ∞)
Решение во вложении...........................