Ответ:
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<span>Xy^2+y+x^2y+x+4xy+4= (xy^2 + x^2y+ 4xy) + (y+x+4) =
=xy(y+x+4) + (y+x+4) = (x+y+4)*(xy+1).</span>
B1=6, q=b2/b1=0,5
B6=b1*q^5 = 6*(0,5^5) =0,1875
<span>5sin^2x+4sin(п/2+x)=4
</span><span>5sin^2x + 4cosx = 4
5*(1 - cos^2x) + 4cosx - 4 = 0
</span>5 - 5cos^2x + 4cosx - 4 = 0
5cos^2x - 4cosx - 1 =
D = 16 + 4*5*1 = 36
<span>cosx = (4 -6)/10
cosx = - (1/5)
x1 = (+ -)arccos(-1/5) + 2</span>πk, k∈Z
<span>cosx = (4 + 6)/10
cosx = 1
x2 = 2</span>πn, n∈Z<span>
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