Y=2sinx x=0 y=0 2sinx=0; snx=0 x=πn n∈Z
y=1-cos(2пи+x)=<span>1-cosx x=0 y=1-1=0
1-cosx=0 cosx=1 x=2</span>πn n∈Z
2х²-4ху²+3ху-6у³=2x(x-2y²)+3y(x-2y²)=(2x+3y)(x-2y²)
При <span>х=1/4,у=1/6
</span>(2*1/4+3*1/6)(1/4-2*(1/6)²)=(1/2+1/2)(1/4-2/36)=1*7/36=7/36
вроде так
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<span>а) ах=а + 6
x = (a+6)/a = 1 + 6/a
б) с(с-2)х=c^2 -4</span>
x = (c^2-4)/c(c-2)
x = (c+2)/c
При с не равном 2