<span>(x-5)(x+5)=x^2-25
(6-y)(6+y)=36-y^2
(a-7)(a+7)=a^2-49
(1-c)(1+c)=1-c^2
(m-4)(m+4)=m^2-16
(n-2)(n+2)=n^2-4
(x-9)(9+×)=x^2-81
(11-d)(d +11)=121-d^2
(4+b)(b-4)=b^2-16
(a+12)(12-a)=144-a^2
(c-3)(3+c)=c^2-9
(8-b)(b+8)=64-b^2
(2x-3)(2x+3)=4x^2-9
(4y-7)(4x+7)=
(8a+5)(5-8a)=25-64a^2
(9c-1)(1+9c)=81c^2-1
(10+3d)(3d-10)=9d^2-100
(11x-6)(6+11x)=121x^2-36
(2m-0,7)(2m+0,7)=4m^2-0.49
(1,2-5y)(1,2+5y)=1.44-25y^2
(8n+0,6)(8n-0,6)=64n^2-0.64
(0,9p+2)(0,9p-2)=0.81p^2-4
(1,1d-8)(8+1,1d)=1.21d^2-64
(0,1a+1)(1-0,1a)=1-0.0.1a^2
(0,7x-3y)(0,7x+3y)=0.49x^2-9y^2
(1,3m-4n)(1,3m+4n)=1.69m^2-16n^2
(5a+0,4d)(0,4d-5a)=0.16d^2-25a^2
(2,5p+6q)(6q-2,5p)=36q^2-6.25p^2
(8c-2,1d)(2,1d+8c)=64c^2-4.41d^2
(0,9x+2y)(0,9x-2y)=0.81x^2-4y^2
(1/4a-0,9b)(1/4a+0,9b)=1/16a^2-0.81b^2
(2/5m-1,5n)(2/5m+1,5n)=4/25m^2-2.25n^2
(3/7x-1/2y)(3/7x+1/2y)=9/49x^2-2/4y^2
(4/9c-9/10d)(9/10d+4/9c)=16/81c^2-81/100d^2
(1/3a+6/7b)(6/7b-1/3a)=36/49b^2-1/9a^2
(1/4a-0,9b)(1/4a+0,9b)=1/16a^2-0.81b^2</span>
2y - x^2 = 1
2y + x^2 = 19
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-x^2 - x^2 = -18
-2x^2 = - 18
2x^2 = 18
x^2 = 9
x1 = 3
x2 = -3
2y = 1 + x^2
2y = 1 + 9
2y = 10
y1 = 5
y2 = 5
Ответ: (3;5) и (-3;5)
6:2=3(км) - половина пути
х км/ч - начальная скорость
(х+10) км/ч - увеличенная скорость
3х+3(х+10)=24*6
6х=144
х=19(км/ч) - с такой скоростью катер плыл первую половину пути
X^{2} +px + q = 0
по теореме Виета
x1 + x2 = -p
x1 * x2 = q
x1 = -5 и x2 = -1
-5 + (-1) = - (-6) =6
-5 * ( -1) =5
значит, q=5