![2x^2+3x+1=x^2-2x-5\\2x^2-x^2+3x+2x+1+5=0\\x^2+5x+6=0\\D=25-4*1*6=1\\x1=\frac{-5+1}{2} =-2\\x2=\frac{-5-1}{2} =-3\\\\](https://tex.z-dn.net/?f=2x%5E2%2B3x%2B1%3Dx%5E2-2x-5%5C%5C2x%5E2-x%5E2%2B3x%2B2x%2B1%2B5%3D0%5C%5Cx%5E2%2B5x%2B6%3D0%5C%5CD%3D25-4%2A1%2A6%3D1%5C%5Cx1%3D%5Cfrac%7B-5%2B1%7D%7B2%7D+%3D-2%5C%5Cx2%3D%5Cfrac%7B-5-1%7D%7B2%7D+%3D-3%5C%5C%5C%5C)
теперь подставляем найденные иксы в любую из функций и находим игрик:
![y(-2)=2*(-2)^2+3(-2)+1=8-6+1=3\\y(-3)=2*(-3)^2+3(-3)+1=18-9+1=10](https://tex.z-dn.net/?f=y%28-2%29%3D2%2A%28-2%29%5E2%2B3%28-2%29%2B1%3D8-6%2B1%3D3%5C%5Cy%28-3%29%3D2%2A%28-3%29%5E2%2B3%28-3%29%2B1%3D18-9%2B1%3D10)
Ответ: точки пересечения (-2;3);(-3;10)
1).
а) =3x/2y.
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б) 5y/(y²-2y) =5y/y(y-2) =5/(y-2) .
в) (3a-3b)/(a²-b²) =3(a-b)/(a-b)(a+b) =3/(a+b).
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2).
а) (3-2a)/2a -(1-a²)/a² =(a(3-2a) -2(1-a²))/2a² =(3a-2a² - 2 +2a²)/2a² = (3a-2)/2a².
* * *3-2a)/2a -(1-a²)/a²=(3/2a -1 -1/a² +1) =3/2a -1/a² = (3a -2)/2a².
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б) (1/(3x+y) - 1/(3x -y) =(3x-y -3x-y)/(3x-y)(3x+y) =-2y/3x-y)(3x+y)
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в) (4-3b)/(b² -2b) +3/(b-2) =(4-3b)/b(b -2) +3/(b-2) =(4-<u>3b</u>+<u>3b</u>)/b(b-2) =4/b(b-2).
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3) (x-6y²)/2y +3y = x/2y -3y +3y =x/2y || x=-8 ;y =0,1 || = -8/2*0,1 = -80/2 =- 40.
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4) 2/(x-4) -(x+8)/(x² -16) -1/x =2/(x-4) -(x+8) / (x -4)(x +4) -1/x =
(2x(x+4) -x(x+8)- (x²+16))/x(x -4)(x +4) =
(2x²+<em>8x</em> -<u>x²</u>- 8x-<u>x²</u>+16)/x(x -4)(x +4) =16/x(x-4)(x+4)..
2ˣ=2x+3 по графику корни примерно х1=3,2 х2=-1,3
cost=-0.6
1)sin²t+cos²t=1
sin²t=1-cos²t=1-(-0.6)²=1-0.36=0.64
sint=+-0.8
так как π/2 ≤ t ≤ π- 2 четверть, следовательно, там sint положительный
значит sint=0.8
2)![tgt=\frac{sint}{cost} =\frac{0.8}{-0.6} =-\frac{8}{6} =-1\frac{1}{3}](https://tex.z-dn.net/?f=tgt%3D%5Cfrac%7Bsint%7D%7Bcost%7D+%3D%5Cfrac%7B0.8%7D%7B-0.6%7D+%3D-%5Cfrac%7B8%7D%7B6%7D+%3D-1%5Cfrac%7B1%7D%7B3%7D)
X²-9≥0 U x≠4
(x-3)(x+3)≥0⇒x=3 U x=-3
+ _ +
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-3 3
x∈(-≈;-3) U (3;4) U (4;≈)