(х-1)(х-5)^2(х-9)=-36
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Ответил mmb1
1
(х-1)(х-5)²(х-9)=-36
(х-1)(х-9)(х-5)²=-36
(x² - 10x + 9)(x² - 10x + 25) = -36
x² - 10x + 17 = y
(y - 8)(y + 8) = -36
y² - 64 = -36
y² = 28
y₁₂ = ±√28
1. x² - 10x + 17 = √28
x² - 10x + 17 - √28 = 0
D = 100 - 4*(17 - √28) = 100 - 68 + 4√28 = 32 + 8√7 = 4(8 + 2√7)
x₁₂ = (10 ± √(32 + 8√7)/2 = 5 ± √(8 + 2√7)
2. x² - 10x + 17 = -√28
x² - 10x + 17 + √28 = 0
D = 100 - 4*(17 + √28) = 100 - 68 - 4√28 = 32 - 8√7 = 4(8 - 2√7)
x₃₄= (10 ± √(32 - 8√7)/2 = 5 ± √(8 - 2√7)
x = { 5 + √(8 + 2√7), 5 - √(8 + 2√7), 5 + √(8 - 2√7), 5 - √(8 - 2√7) }
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но вероятнее
(x² - 10x + 9)(x² - 10x + 25) = 36
тогда
y12 = +- 10
1. x² - 10x + 17 = 10
x² - 10x + 7 = 0
D = 100 - 28 = 72
x₁₂ = (10 ± √72)/2 = 5 +- 3√2
2. x² - 10x + 17 = -10
x² - 10x + 27 = 0
D = 100 - 4*27 < 0
x = { 5 +- 3√2}
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