решите дробно-рациональные уравнения
1) 1/x-5 - 1/x = 1, в ответе укажите сумму корней.
2) x -2= 4x-14/x-3
3)4/x^2 +4 + 5 / x^2 - 5 =2
4) 3- 7x/ 2x+4 = 1,5 - 3,5x / x+2
5) x/x+1 + x+1/x+2 + x+2/x =3
Ответы
Ответ дал:
0
1) ![frac{1}{x-5} - frac{1}{x}=1 frac{1}{x-5} - frac{1}{x}=1](https://tex.z-dn.net/?f=+frac%7B1%7D%7Bx-5%7D+-+frac%7B1%7D%7Bx%7D%3D1+)
Общ. знам.: x (x-5) ≠0
x - (x-5) = x(x-5)
x - x + 5 = x² - 5x
x² - 5x - 5 = 0
D = 25 + 20 = 45
x₁ =![frac{5+ sqrt{45} }{2} frac{5+ sqrt{45} }{2}](https://tex.z-dn.net/?f=+frac%7B5%2B+sqrt%7B45%7D+%7D%7B2%7D+)
x₂=![frac{5- sqrt{45} }{2} frac{5- sqrt{45} }{2}](https://tex.z-dn.net/?f=+frac%7B5-+sqrt%7B45%7D+%7D%7B2%7D+)
x₁ + x₂ =![frac{5+ sqrt{45} }{2} + frac{5- sqrt{45} }{2} = frac{5+ sqrt{45}+5- sqrt{45} }{2} = frac{10}{2} =5 frac{5+ sqrt{45} }{2} + frac{5- sqrt{45} }{2} = frac{5+ sqrt{45}+5- sqrt{45} }{2} = frac{10}{2} =5](https://tex.z-dn.net/?f=+frac%7B5%2B+sqrt%7B45%7D+%7D%7B2%7D+%2B+frac%7B5-+sqrt%7B45%7D+%7D%7B2%7D+%3D+frac%7B5%2B+sqrt%7B45%7D%2B5-+sqrt%7B45%7D++%7D%7B2%7D+%3D+frac%7B10%7D%7B2%7D+%3D5)
Ответ: 5
2)![x-2= frac{4x-14}{x-3} x-2= frac{4x-14}{x-3}](https://tex.z-dn.net/?f=x-2%3D+frac%7B4x-14%7D%7Bx-3%7D+)
Общ. знам.: x-3≠0
(x-2)(x-3) = 4x-14
x² - 3x - 2x + 6 = 4x - 14
x² - 9x + 20 = 0
D = 81-80=1
x₁ = (9+1) /2 = 5
x₂ = (9-1) / 2 = 4
Ответ: 5; 4
3)![frac{4}{ x^{2} +4} + frac{5}{ x^{2} -5}=2 frac{4}{ x^{2} +4} + frac{5}{ x^{2} -5}=2](https://tex.z-dn.net/?f=+frac%7B4%7D%7B+x%5E%7B2%7D+%2B4%7D+%2B+frac%7B5%7D%7B+x%5E%7B2%7D+-5%7D%3D2+)
Общ. знам.: (x² + 4)(x² - 5) ≠ 0
4(x²-5) + 5(x²+ 4) = 2(x² +4) (x²-5)
4x² - 20 + 5x² + 20 = 2x⁴ - 10x² + 8x² - 40
2x⁴ -11x² - 40 = 0
x² = t; t≥0
2t² - 11t - 40 = 0
D = 121 + 320 = 441
t₁ = (11+21) / 4 = 8
t₂ = (11-21) / 4 = -2.5 не удов. усл.: t≥0
x² = 8
x₁ =![sqrt{8}=2 sqrt{2} sqrt{8}=2 sqrt{2}](https://tex.z-dn.net/?f=+sqrt%7B8%7D%3D2+sqrt%7B2%7D++)
x₂ =![-2 sqrt{2} -2 sqrt{2}](https://tex.z-dn.net/?f=-2+sqrt%7B2%7D+)
Ответ:![2 sqrt{2} ;-2 sqrt{2} 2 sqrt{2} ;-2 sqrt{2}](https://tex.z-dn.net/?f=2+sqrt%7B2%7D+%3B-2+sqrt%7B2%7D+)
4)![frac{3-7x}{2x+4}= frac{1.5-3.5x}{x+2} frac{3-7x}{2x+4}= frac{1.5-3.5x}{x+2}](https://tex.z-dn.net/?f=+frac%7B3-7x%7D%7B2x%2B4%7D%3D+frac%7B1.5-3.5x%7D%7Bx%2B2%7D++)
![frac{3-7x}{2(x+2)} = frac{1.5-3.5x}{x+2} frac{3-7x}{2(x+2)} = frac{1.5-3.5x}{x+2}](https://tex.z-dn.net/?f=+frac%7B3-7x%7D%7B2%28x%2B2%29%7D+%3D+frac%7B1.5-3.5x%7D%7Bx%2B2%7D+)
Общ. знам.: 2(x+2) ≠ 0
3-7x = 2*(1.5-3.5x)
3 - 7x = 3 - 7x
0 = 0
x - любое, НО 2(x+2)≠0 ⇒ x≠-2
x∈ (-беск.; -2) U (-2; +беск.)
Ответ: (-беск.; -2) U (-2; +беск.)
5)![frac{x}{x+1}+ frac{x+1}{x+2}+ frac{x+2}{x}=3 frac{x}{x+1}+ frac{x+1}{x+2}+ frac{x+2}{x}=3](https://tex.z-dn.net/?f=+frac%7Bx%7D%7Bx%2B1%7D%2B+frac%7Bx%2B1%7D%7Bx%2B2%7D%2B+frac%7Bx%2B2%7D%7Bx%7D%3D3+++)
Общ. знам.: x (x+1)(x+2) ≠ 0
x*x(x+2) + (x+1)(x+1)*x + (x+2)*(x+1)*(x+2) = 3x(x+1)(x+2)
x³ + 2x² + x*(x+1)² + (x+1)(x+2)² = 3x (x² + 2x + x + 2)
x³ + 2x² + x (x² + 2x + 1) + (x+1)(x²+4x+4) = 3x³ + 9x² + 6x
x³ + 2x² +x³ + 2x² +x + x³ + 4x² + 4x + x² + 4x+ 4 = 3x³ + 9x² + 6x
3x³ + 9x² +9x + 4 = 3x³ + 9x² + 6x
3x = -4
x =![- frac{4}{3} =-1 frac{1}{3} - frac{4}{3} =-1 frac{1}{3}](https://tex.z-dn.net/?f=-+frac%7B4%7D%7B3%7D+%3D-1+frac%7B1%7D%7B3%7D+)
Ответ:
Общ. знам.: x (x-5) ≠0
x - (x-5) = x(x-5)
x - x + 5 = x² - 5x
x² - 5x - 5 = 0
D = 25 + 20 = 45
x₁ =
x₂=
x₁ + x₂ =
Ответ: 5
2)
Общ. знам.: x-3≠0
(x-2)(x-3) = 4x-14
x² - 3x - 2x + 6 = 4x - 14
x² - 9x + 20 = 0
D = 81-80=1
x₁ = (9+1) /2 = 5
x₂ = (9-1) / 2 = 4
Ответ: 5; 4
3)
Общ. знам.: (x² + 4)(x² - 5) ≠ 0
4(x²-5) + 5(x²+ 4) = 2(x² +4) (x²-5)
4x² - 20 + 5x² + 20 = 2x⁴ - 10x² + 8x² - 40
2x⁴ -11x² - 40 = 0
x² = t; t≥0
2t² - 11t - 40 = 0
D = 121 + 320 = 441
t₁ = (11+21) / 4 = 8
t₂ = (11-21) / 4 = -2.5 не удов. усл.: t≥0
x² = 8
x₁ =
x₂ =
Ответ:
4)
Общ. знам.: 2(x+2) ≠ 0
3-7x = 2*(1.5-3.5x)
3 - 7x = 3 - 7x
0 = 0
x - любое, НО 2(x+2)≠0 ⇒ x≠-2
x∈ (-беск.; -2) U (-2; +беск.)
Ответ: (-беск.; -2) U (-2; +беск.)
5)
Общ. знам.: x (x+1)(x+2) ≠ 0
x*x(x+2) + (x+1)(x+1)*x + (x+2)*(x+1)*(x+2) = 3x(x+1)(x+2)
x³ + 2x² + x*(x+1)² + (x+1)(x+2)² = 3x (x² + 2x + x + 2)
x³ + 2x² + x (x² + 2x + 1) + (x+1)(x²+4x+4) = 3x³ + 9x² + 6x
x³ + 2x² +x³ + 2x² +x + x³ + 4x² + 4x + x² + 4x+ 4 = 3x³ + 9x² + 6x
3x³ + 9x² +9x + 4 = 3x³ + 9x² + 6x
3x = -4
x =
Ответ:
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