1
(ab+ac)²/(ab²-ac²)=a²(b+c)/a(b²-c²)=a(b+c)/(b-c)(b+c)=a/(b-c)
2
1)1/c-1/(10-c)=(10-c-c)/c(10-c)=(10-2c)/c(10-c)=2(5-c)/c(10-c)=2(c-5)/c(c-10)
2)25/(c-5) -(c-5)=(25-c²+10c-25)/(c-5)=(10c-c²)/(c-5)=c(10-c)/(c-5)
3)2(c-5)/c(c-10) * c(10-c)/(c-5)=-2
X + y = z + 1
z - y = 3
X = 2y
Решение
2у + у = z + 1
3y = z + 1 ==> z = 3y - 1
z - y = 3
3y - 1 - y = 3
2y = 4
y = 2
X = 2 • 2 = 4
Z = 3 + y
Z = 3 + 2 = 5
Ответ ( 4 ; 2 ; 5 )
Х+у=-3 | 3
3х+3у= -9
(3х+3у)+(-3х+3у)=-18
6у=-18
у=- 18:6
у=-3
х+у=-3
х=3+(-3)
х=0
Сделала решение на листочке
<span><span><span>Применим формулу замены произведения косинусов их суммой
cos α · cos β</span>
=
(cos(α-β) + cos(α+β))/
</span><span>
2</span></span>
cos44*cos16=1/2(cos60+cos28)=1/2(1/2+cos28), cos60=1/2
cos 59*cos31=1/2(cos90+cos28)=1/2<span>cos28, </span> cos90=0
1/2(1/2+cos28)-1/2<span>cos28=1/4+1/2</span>cos28-1/2cos28=1/4