<span>2sinxcosx=sinx/cosx</span>
<span><span>tgx(2cos^2x -1)=0</span></span>
<span><span><span>tgx=0 не уд одз </span></span></span>
<span><span><span><span>cosx= sqrt(1/2)</span></span></span></span>
<span><span><span><span><span>x=Pi/2+Pn</span></span></span></span></span>
9(х∧2)∧2+23х∧2-12=0
x∧2=a
9a∧2+23a-12=0
Д=b∧2-4ac=23∧2-4*9*(-12)=529+432=961
=
=
=
=
=
=
=
=
=
= -3
x∧2=
x=
x=
=плюсминус
x∧2=-3
x
Ответ:
; -
1)
5(x-y)²+(x-2y)²=5(x²-2xy+y²)+x²-4xy+4y²=
=5x²-10xy+5y²+x²-4xy+4y²=6x²-14xy+9y²
2)
4(m-2n)²-3(3m-n)²=4(m²-4mn+4n²)-3(9m²-6mn+n²)=
=4m²-16mn+16n²-27m²+18mn-3n²=-23m²+2mn+13n²
3)
(2a-b)²-5(a-2b)²=4a²-4ab+b²-5(a²-4ab+4b²)=4a²-4ab+b²-5a²+20ab-20b²=
=-a²+16ab-19b²
4)
(3x+4y)²-7(2x-3y)²=9x²+24xy+16y²-7(4x²-12xy+9y²)=
=9x²+24xy+16y²-28x²+84xy-63y²=-19x²+108xy-47y²
5)
2(p-3g)²-4(2p-g)²-(2g-3p)(p+g)=
=2(p²-6pg+9g²)-4(4p²-4pg+g²)--(2pg+2g²-3p²-3pg)=
=2p²-12pg+18g²-16p²+16pg-4g²-2pg-2g²+3p²+3pg=
=-11p²+5pg+12g²
6)
5(n-5m)²-6(2m-3n)²-(3m-n)(7m-n)=
=5(n²-10mn+25m²)-6(4m²-12mn+9n²)-(21m²-3mn-7mn+n²)=
=5n²-50mn+125m²-24m²+72mn-54n²-21m²+3mn+7mn-n²=
=-50n²+32mn+80m²
7)
(x-y)³=x³-2x²y+2y²x-y³
8)
(2a-b)³=8a³-8a²b+2ab²-b³
Нужно разделить обе части уравнения на cosx, и мы получим tgx + 1 = 0, или tgx = -1. Отсюда получаем x = -p/4 + pn
Пусть √a=t, a √y=n, тогда
=(t²-n²)/(t+n)=[(t-n)(t+n)]/(t+n)=t-n
t-n=√a-√y
