3x+2y+7z=3x+y+y+7z=3x+y + 7z+y= 8+6=14
Ответ: 14
<span>x²-7x+12≥0
(x-3)(x-4)≥0
x1=3
x2=4
_____+____3_____-______4____+_____x
Ответ: x∈(-∞;3] и [4;+∞)
</span>
1) =5(x-3)-3(x-3)^2=(x-3)(5-3x+9)=(x-3)(14-3x)
2) =0,73(0,73+0,27)+0,27=0,73+0,27=1
3)12^5-18^4=2^5*6^5-3^4*6^4=6^4(32*6-81)=6^4(192-81)=6^4*111,
б) =6^4*111=6^3*666, (666:37)
4)2x^2-x=0, x(2x-1)=0, x=0, x=1/2
5(x-3)-(6-2x)^2=0, 5(x-3)-(2(3-x)^2=0, (x-3)(5-4(x-3))=0, (x-3)(8-4x)=0, x=3, x=2
Х^6 - х^2 * у^4 = х^6 - х^2у^4 = х^2*(х^4 - у^4) = х^2*(х^2 - у^2)(х^2 + у^2) = х^2*(х - у)*(х+у)*(х^2 + у^2).