Ответ:
6. x = 5.
7. x₁ = 1, x₂ = 5.
8. x = 1.
9. x ∉ R.
10. x ∉ ∅.
11. x = 1.
12. x = 4.
Пошаговое объяснение:
6.
x - 1 = 4
x = 5
Проверка:
3 = 3.
7.
(x - 1) * (4 - 1 (x - 1)) = 0
(x - 1) * (4 - x + 1) = 0
(x - 1) * (5 - x) = 0
x₁ = 1, x₂ = 5
8.
D = b² - 4ac = 9 + 16 = 25 ()
x₁ =
x₂ =
x = 1.
9.
x ∈ R
10.
x ∈ ∅
11.
D = b² - 4ac = 4 + 60 = 64 ()
12.
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Answers & Comments
Ответ:
6. x = 5.
7. x₁ = 1, x₂ = 5.
8. x = 1.
9. x ∉ R.
10. x ∉ ∅.
11. x = 1.
12. x = 4.
Пошаговое объяснение:
6.
x - 1 = 4
x = 5
Проверка:
3 = 3.
7.
(x - 1) * (4 - 1 (x - 1)) = 0
(x - 1) * (4 - x + 1) = 0
(x - 1) * (5 - x) = 0
x₁ = 1, x₂ = 5
8.
D = b² - 4ac = 9 + 16 = 25 (
)
x₁ =
x₂ =
x = 1.
9.
x ∈ R
10.
x ∈ ∅
11.
D = b² - 4ac = 4 + 60 = 64 (
)
x₁ =![\frac{-b+\sqrt{D} }{2a} = \frac{-2+8}{6} = 1 [/tex}x₂ = [tex]\frac{-b-\sqrt{D} }{2a} = \frac{-2-8}{6} = \frac{5}{3} [/tex}[tex]\left \{ {\sqrt{1+1}+\sqrt{1-1}=\sqrt{3*1-1}} \atop {\sqrt{-\frac{5}{3}+1}}}+\sqrt{-\frac{5}{3} - 1} = \sqrt{3*(-\frac{5}{3}) -1} \right. \left \{ {{1,41421=1,41421} \atop {\sqrt{-\frac{2}{3}}}+\sqrt{-\frac{5}{3} - 1} = \sqrt{3*(-\frac{5}{3}) -1}} \right. \left \{ {{x=1} \atop {x\neq -\frac{5}{3}}} \right. \frac{-b+\sqrt{D} }{2a} = \frac{-2+8}{6} = 1 [/tex}x₂ = [tex]\frac{-b-\sqrt{D} }{2a} = \frac{-2-8}{6} = \frac{5}{3} [/tex}[tex]\left \{ {\sqrt{1+1}+\sqrt{1-1}=\sqrt{3*1-1}} \atop {\sqrt{-\frac{5}{3}+1}}}+\sqrt{-\frac{5}{3} - 1} = \sqrt{3*(-\frac{5}{3}) -1} \right. \left \{ {{1,41421=1,41421} \atop {\sqrt{-\frac{2}{3}}}+\sqrt{-\frac{5}{3} - 1} = \sqrt{3*(-\frac{5}{3}) -1}} \right. \left \{ {{x=1} \atop {x\neq -\frac{5}{3}}} \right.](https://tex.z-dn.net/?f=%5Cfrac%7B-b%2B%5Csqrt%7BD%7D%20%7D%7B2a%7D%20%3D%20%5Cfrac%7B-2%2B8%7D%7B6%7D%20%3D%201%20%20%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3Ex%E2%82%82%20%3D%20%5Btex%5D%5Cfrac%7B-b-%5Csqrt%7BD%7D%20%7D%7B2a%7D%20%3D%20%5Cfrac%7B-2-8%7D%7B6%7D%20%3D%20%5Cfrac%7B5%7D%7B3%7D%20%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cleft%20%5C%7B%20%7B%5Csqrt%7B1%2B1%7D%2B%5Csqrt%7B1-1%7D%3D%5Csqrt%7B3%2A1-1%7D%7D%20%5Catop%20%7B%5Csqrt%7B-%5Cfrac%7B5%7D%7B3%7D%2B1%7D%7D%7D%2B%5Csqrt%7B-%5Cfrac%7B5%7D%7B3%7D%20-%201%7D%20%3D%20%5Csqrt%7B3%2A%28-%5Cfrac%7B5%7D%7B3%7D%29%20-1%7D%20%5Cright.%20%5Cleft%20%5C%7B%20%7B%7B1%2C41421%3D1%2C41421%7D%20%5Catop%20%7B%5Csqrt%7B-%5Cfrac%7B2%7D%7B3%7D%7D%7D%2B%5Csqrt%7B-%5Cfrac%7B5%7D%7B3%7D%20-%201%7D%20%3D%20%5Csqrt%7B3%2A%28-%5Cfrac%7B5%7D%7B3%7D%29%20-1%7D%7D%20%5Cright.%20%5Cleft%20%5C%7B%20%7B%7Bx%3D1%7D%20%5Catop%20%7Bx%5Cneq%20-%5Cfrac%7B5%7D%7B3%7D%7D%7D%20%5Cright.)
12.
Проверка: