4(3-x)=6(x+1)-15x
12-4x=6x+6-15x
12-4x=-9x+6
-4x+9x=6x-12
5x=-6
x=-6/5
Log(127+x³)-3*log(x+1)=0
ОДЗ:
![\left \{ {{127+ x^{3} \ \textgreater \ 0} \atop {x+1\ \textgreater \ 0}} \right. , \left \{ {{ x^{3}\ \textgreater \ -127 } \atop {x\ \textgreater \ -1}} \right. , \left \{ {{x\ \textgreater \ \sqrt[3]{-127} } \atop {x\ \textgreater \ -1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B127%2B+x%5E%7B3%7D+%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7Bx%2B1%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%2C++++%5Cleft+%5C%7B+%7B%7B+x%5E%7B3%7D%5C+%5Ctextgreater+%5C+-127+%7D+%5Catop+%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D+%5Cright.+%2C++++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C++%5Csqrt%5B3%5D%7B-127%7D+%7D+%5Catop+%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D+%5Cright.+)
x>-1. x∈(-1;∞)
log(127+x³)=3*log(x+1)
log(127+x³)=log(x+1)³
127+x³=(x+1)³
127+x³=x³+3x²+3x+1
3x²+3x-126=0 |:3
x²+x-42=0
D=169
x₁=-7, x₂=6
-7∉(-1;∞). => x=-7 посторонний корень
ответ: x=6
4х - 3у = -1
х - 5у = 4
4х - 3у = -1
х = 4 + 5у
4 (4 + 5у) - 3у = -1
х = 4 + 5у
16 + 20у - 3у = -1
х = 4 + 5у
17у = -17
х = 4 + 5у
у = -1
х = 4 + 5* (-1)
у = 1
х = 4 - 5
у = 1
х = -1
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