1) (2n²-3n+2n-3)+(3n²+n-3n-1)=5n²-3n-4
2) (2x²-3xy-2xy+3y²)-(6x²+3xy-2xy-y²) =4y²-4x²-6xy=2(2y²-2x²-3xy)
3) (4a²+6a+6a+9)-(4a²-2a+2a-1)=12a+10=2(6a+5)
4) (3cd+9c²-d²-3cd)+(4c²-16cd-cd+4d²) =13c²-17cd+d²
Solve for x over the real numbers:
x^2+4 = 6-x
Subtract 6-x from both sides:
x^2+x-2 = 0
The left hand side factors into a product with two terms:
(x-1) (x+2) = 0
Split into two equations:
x-1 = 0 or x+2 = 0
Add 1 to both sides:
x = 1 or x+2 = 0
Subtract 2 from both sides:
Answer: |
| x = 1 or x = -2
Удачи!
А) По свойствам логарифма
log3 (sin^2 x) = 2*log3 (sin x)
Сделаем замену t = log3 (sin x)
t^2 + 2t = log3(2)*t
t^2 + t*(2 - log3(2) ) = 0
t*(t + 2 - log3(2) ) = 0
1) t = log3 (sin x) = 0
sin x = 1
x1 = pi/2 + 2pi*n
2) t = log3(2) - 2
log3 (sin x) = log3(2) - log3(9) = log3(2/9)
sin x = 2/9
x2 = arcsin(2/9) + 2pi*k
x3 = pi - arcsin(2/9) + 2pi*k
Б) arcsin(2/9)≈2/9=0,22 < pi/3, поэтому в [pi/3; 2pi] попадают корни:
x1 = pi/2; x2 = pi - arcsin(2/9)
<span>Против наибольшего угла всегда лежит наибольшая сторона, т.е. угол будет между сторонами 5 и 6</span>
теперь воспользуемся теоремой косинусов
7^2=5^2+6^2-2*5*6*cosA
49=25+36-60*cosA
cosA=0.2