1) Cosx = t
6t² + t -1 = 0
D = b² -4ac = 1 - 4*6*(-1) = 25 > 0
t₁ = (-1+5)/12 = 4/12 = 1/3
t₂ = (-1 -5)/12 = -1/2
a) Cosx = 1/3 б) Сosx = -1/2
x = +-arcCos(1/3) + 2πk , k ∈Z x = +-arcCos(-1/2) + 2πn , n ∈Z
x = +- 2π/3 +2πn , n ∈ Z
2) учтём, что Cosx = 2Cos²x/2 -1
наше уравнение:
Cosx/2 = 1 + 2Cos²x/2 -1
Cosx/2 = t
2Cos²x/2 - Cosx/2 = 0
Cosx/2(2Cosx/2 -1) = 0
Cosx/2 = 0 или 2Cosx/2 -1 = 0
x/2 = π/2 + 2πk , k ∈Z Cosx/2 = 1/2
x = π + 4πk , k ∈ Z x/2 = +-arcCos(1/2) + 2πn , n ∈ Z
x/2= +- π/3+ 2πn , n ∈ Z
x = +-2π/3 + 4 πn , n ∈ Z
1) Cosx = t
6t² + t -1 = 0
D = b² -4ac = 1 - 4*6*(-1) = 25 > 0
t₁ = (-1+5)/12 = 4/12 = 1/3
t₂ = (-1 -5)/12 = -1/2
a) Cosx = 1/3 б) Сosx = -1/2
x = +-arcCos(1/3) + 2πk , k ∈Z x = +-arcCos(-1/2) + 2πn , n ∈Z
x = +- 2π/3 +2πn , n ∈ Z
2) учтём, что Cosx = 2Cos²x/2 -1
наше уравнение:
Cosx/2 = 1 + 2Cos²x/2 -1
Cosx/2 = t
2Cos²x/2 - Cosx/2 = 0
Cosx/2(2Cosx/2 -1) = 0
Cosx/2 = 0 или 2Cosx/2 -1 = 0
x/2 = π/2 + 2πk , k ∈Z Cosx/2 = 1/2
x = π + 4πk , k ∈ Z x/2 = +-arcCos(1/2) + 2πn , n ∈ Z
x/2= +- π/3+ 2πn , n ∈ Z
x = +-2π/3 + 4 πn , n ∈ Z
1) y1=-6; y2=5
2) x1=7/5; x2=3
Объяснение:
1)![\frac{2y-2}{y+3}+\frac{y+3}{y-3} =5\\\\ \frac{2y^{2}-8y+6+y^{2}+6y+9-5y^{2} +45}{(y+3)(y-3)} =0\\ \\ \frac{-2y^{2}-2y+60}{(y+3)(y-3)} =0\\\\ \frac{y^{2}+y-30}{(y+3)(y-3)} =0\\\\ (y+6)(y-5)=0\\y=-6; y=5\\(y+3)(y-3) \neq 0\\y\neq 3; y\neq -3](/tpl/images/1018/7053/9e753.png)
2)![\frac{1}{x-1} +\frac{1}{x^{2} -1} =\frac{5}{8} \\\\ \frac{8x+8+8-5x^{2}+5 }{8(x+1)(x-1)} =0\\\\ \frac{-5x^{2}+8x+21 }{8(x+1)(x-1)} =0\\ \\x\neq1; x\neq -1\\ 5x^{2} -8x-21=0\\(5x+7)(x-3)=0\\x1=-\frac{7}{5}; x2=3](/tpl/images/1018/7053/cb803.png)