(x-|y|=2
(4x+3|y|=15
(x=2+|y|
(4x+3|y|=15
4(2+|y|)+3×|y|=15
y=1
y=-1
x=2+|1|
x=2+|-1|
x=3
x=3
(x1y1)=(3,1)
(x2y2)=(3,-3)
(3-|1|=2
(4×3+3×|1|=15
(3-|-1|=2
(4×3+3×|-1|=15
(2=2
(15=15
(2=2
(15=15
ответ:
(x1y1)=(3,1)
(x2y2)=(3,-1)
Tg 3x=1/√3;
3x=atctg (1/√3)+πk, k∈Z;
3x=(π/6)+πk, k∈Z;
x=(π/18) + (π/3)k, k∈Z.
О т в е т. (π/18) + (π/3)k, k∈Z.
Пусть х это-скорость первой машины. Тогда х+10 равна- скорость второй машины.
Составим уравнение-—
2,5хравен2(х+10)
2,5хравен2х+20
0,5хравен20
хравен40 км/ч скорость первой машины
40+10равен50 км/ч скорость второй машины.
3(mn)-1y= 3 х mn-1y= 3mn-1y
![a^3-b^3=(a-b)(a^2+ab+b^2)](https://tex.z-dn.net/?f=a%5E3-b%5E3%3D%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29)
1)
![(2x+y)^3-x^3=(2x+y-x)(4x^2+2xy+y^2)= \\ \\ =(x+y)(4x^2+2xy+y^2)](https://tex.z-dn.net/?f=%282x%2By%29%5E3-x%5E3%3D%282x%2By-x%29%284x%5E2%2B2xy%2By%5E2%29%3D+%5C%5C++%5C%5C+%3D%28x%2By%29%284x%5E2%2B2xy%2By%5E2%29)
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2)
![8p^3-(p+1)^3=(2p)^3-(p+1)^3= \\ \\ =(2p-(p+1))(4p^2+2p(p+1)+(p+1)^2)= \\ \\ =(p+1)(4p^2+2p^2+2p+p^2+2p+1)= \\ \\ =(p+1)(7p^2+4p+1)](https://tex.z-dn.net/?f=8p%5E3-%28p%2B1%29%5E3%3D%282p%29%5E3-%28p%2B1%29%5E3%3D+%5C%5C++%5C%5C+%3D%282p-%28p%2B1%29%29%284p%5E2%2B2p%28p%2B1%29%2B%28p%2B1%29%5E2%29%3D+%5C%5C++%5C%5C+%3D%28p%2B1%29%284p%5E2%2B2p%5E2%2B2p%2Bp%5E2%2B2p%2B1%29%3D+%5C%5C++%5C%5C+%3D%28p%2B1%29%287p%5E2%2B4p%2B1%29)