1) sinα = 0,6; α ∈ (π/2; π) ⇒ cosα = √1-sin²α ; cosα = - √1-0,36 = - 0,8 ⇒
- 3sin(5π/2 - α) = -3 cosα = -3 · ( - 0,8) = 2,4
Ответ: - 3sin(5π/2 - α) = 2,4
![1.~ a)~ (x+4)^2=x^2+8x+16\\ b)~ (y-5x)^2=y^2-10xy+25y^2\\ c)~ (3a-2)(3a+2)=(3a)^2-2^2=9a^2-4\\ d)~ (c-2b)(c+2b)=c^2-(2b)^2=c^2-4b^2](https://tex.z-dn.net/?f=1.~+a%29~+%28x%2B4%29%5E2%3Dx%5E2%2B8x%2B16%5C%5C+b%29~+%28y-5x%29%5E2%3Dy%5E2-10xy%2B25y%5E2%5C%5C+c%29~+%283a-2%29%283a%2B2%29%3D%283a%29%5E2-2%5E2%3D9a%5E2-4%5C%5C+d%29~+%28c-2b%29%28c%2B2b%29%3Dc%5E2-%282b%29%5E2%3Dc%5E2-4b%5E2)
2. Разложить на множители:
![a)~ x^2-81=x^2-9^2=(x-9)(x+9)\\ b)~ y^2-4y+4=(y-2)^2](https://tex.z-dn.net/?f=a%29~+x%5E2-81%3Dx%5E2-9%5E2%3D%28x-9%29%28x%2B9%29%5C%5C+b%29~+y%5E2-4y%2B4%3D%28y-2%29%5E2)
в пункте б) опечатка, так что предположил как должно быть
![c)~ 36x^4y^2-169c^2=(6x^2y)^2-(13c)^2=(6x^2y-13c)(6x^2y+13c)\\ d)~ (x+1)^2-(x-1)^2=(x+1-x+1)(x+1+x-1)=2\cdot 2x=4x](https://tex.z-dn.net/?f=c%29~+36x%5E4y%5E2-169c%5E2%3D%286x%5E2y%29%5E2-%2813c%29%5E2%3D%286x%5E2y-13c%29%286x%5E2y%2B13c%29%5C%5C+d%29~+%28x%2B1%29%5E2-%28x-1%29%5E2%3D%28x%2B1-x%2B1%29%28x%2B1%2Bx-1%29%3D2%5Ccdot+2x%3D4x)
3. Упростить выражение:
![(c+6)^2-c(c+12)=c^2+12c+36-c^2-12c=36](https://tex.z-dn.net/?f=%28c%2B6%29%5E2-c%28c%2B12%29%3Dc%5E2%2B12c%2B36-c%5E2-12c%3D36)
4. Решите уравнение:
![a)~ (x+7)^2-(x-4)(x+4)=65\\ x^2+14x+49-x^2+16=65\\ 14x=0\\ x=0](https://tex.z-dn.net/?f=a%29~+%28x%2B7%29%5E2-%28x-4%29%28x%2B4%29%3D65%5C%5C+x%5E2%2B14x%2B49-x%5E2%2B16%3D65%5C%5C+14x%3D0%5C%5C+x%3D0)
![b)~ 49y^2-64=0\\ y^2=\dfrac{64}{49}~~\Rightarrow~~~ y_{1,2}=\pm\dfrac{8}{7}](https://tex.z-dn.net/?f=b%29~+49y%5E2-64%3D0%5C%5C+y%5E2%3D%5Cdfrac%7B64%7D%7B49%7D~~%5CRightarrow~~~+y_%7B1%2C2%7D%3D%5Cpm%5Cdfrac%7B8%7D%7B7%7D)
5. Выполнить действия:
![a)~ (4a^2+b^2)(2a-b)(2a+b)=(4a^2+b^2)(4a^2-b^2)=16a^4-b^4\\ b)~ (b^2c^3-2a^2)(b^2c^3+2a^2)=(b^2c^3)^2-(2a^2)^2=b^4c^6-4a^4](https://tex.z-dn.net/?f=a%29~+%284a%5E2%2Bb%5E2%29%282a-b%29%282a%2Bb%29%3D%284a%5E2%2Bb%5E2%29%284a%5E2-b%5E2%29%3D16a%5E4-b%5E4%5C%5C+b%29~+%28b%5E2c%5E3-2a%5E2%29%28b%5E2c%5E3%2B2a%5E2%29%3D%28b%5E2c%5E3%29%5E2-%282a%5E2%29%5E2%3Db%5E4c%5E6-4a%5E4)
6*.Докажите неравенство:
![4x^2+9y^2>12xy-0.1\\ 4x^2-12xy+9y^2>-0.1\\ (2x-3y)^2>-0.1](https://tex.z-dn.net/?f=4x%5E2%2B9y%5E2%3E12xy-0.1%5C%5C+4x%5E2-12xy%2B9y%5E2%3E-0.1%5C%5C+%282x-3y%29%5E2%3E-0.1)
Что и требовалось доказать
Ответ:
при t>-1 кроме t=0
Объяснение:D=4+4t>0 , 4+4t>0, 4t>-4, t>-1. Если t=0, то 2х-1=0, х=1/2 (0,5)