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colak
@colak
July 2022
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Решите неравенство:(х^2-4х)^2≥
16
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Verified answer
(x^2 - 4x)^2 ≥16;
(x^2 - 4x)= t ;
t^2 - 4^2 ≥ 0;
(t-4)(t+4) ≥0;
+ - +
_______(-4)________(4)____t
t ≤ - 4 U t≥4;
1) t ≤ -4; ⇒ x^2 - 4x ≤ -4;
x^2 - 4x +4 ≤0;
(x-2)^2 ≤ 0;
x = 2.
2) t ≥ 4; ⇒x^2 - 4x ≥ 4;
x^2 - 4x - 4 ≥ 0;
D = 16 +16 = 32 = (4sgrt2)^2;
x1 = (4+4sgrt2) /2= 2 +2 sgrt2;
x2 = 2 - 2sgrt2.
+ - +
____(2-2sgrt2)_________(2+2sgrt2)_______x
x∈(- беск-сть; 2- 2sgrt2] U [2+2sgrt2; + беск-ность)
Объединим получившиеся решения и получим ответ:
х ∈( - беск-сть;2 -2sgrt2] U {2} U ( 2+2sgrt2; +беск-сть).
6 votes
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Answers & Comments
Verified answer
(x^2 - 4x)^2 ≥16;(x^2 - 4x)= t ;
t^2 - 4^2 ≥ 0;
(t-4)(t+4) ≥0;
+ - +
_______(-4)________(4)____t
t ≤ - 4 U t≥4;
1) t ≤ -4; ⇒ x^2 - 4x ≤ -4;
x^2 - 4x +4 ≤0;
(x-2)^2 ≤ 0;
x = 2.
2) t ≥ 4; ⇒x^2 - 4x ≥ 4;
x^2 - 4x - 4 ≥ 0;
D = 16 +16 = 32 = (4sgrt2)^2;
x1 = (4+4sgrt2) /2= 2 +2 sgrt2;
x2 = 2 - 2sgrt2.
+ - +
____(2-2sgrt2)_________(2+2sgrt2)_______x
x∈(- беск-сть; 2- 2sgrt2] U [2+2sgrt2; + беск-ность)
Объединим получившиеся решения и получим ответ:
х ∈( - беск-сть;2 -2sgrt2] U {2} U ( 2+2sgrt2; +беск-сть).