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KsenijaVolk84
@KsenijaVolk84
October 2021
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Помогите решить неравенства ,пожалуйста
1)x^4-4x^3+4x^2-4 больше или равно 0
2) (2x^2-x)^2<1
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Verified answer
1)
x^4 -4x³ +4x² -4 ≥ 0;
x²(x² -4x +4) -4 ≥ 0;
(x(x-2))² -2² ≥ 0 ;
(x(x-2) -2)(x(x-2) + 2) ≥0;
(x² -2x -2)(x² -2x +2)≥ 0;
т.к. x² -2x+2 =(x-1)² +1 ≥1 >0,
то x² -2x -2 ≥ 0 ; [ (x-x₁)(x-x₂) ≥ 0 ] ***
( x -1 + √5)(x -1 -√5) ≥ 0 ;
x∈
( - ∞ ; 1-√5] U [ 1+√5 ; ∞) .
===================================
+ - +
----------------- 1 -√5 ---------- 1+√5 ------------
================================
2)
(2x² -x)² <1;
(2x² -x)² -1 <0 ;
(2x² -x -1)(2x² - x +1) <0 ;
т.к. 2x² - x +1 =2(x-1/4)² +7/8 ≥7/8 >0;
[
D =1² -4*2*1 = -7 <0; 2>0 ⇒ 2x² - x +1 >0.
]
то 2x² -x -1 <0;
2(x+1/2)(x-1) <0 ;
x∈ (-1/2 ; 1).
===============================
2x² -x -1 = 0 ;
D =1²-4*2(-1) =9 =3² ;
x =(1+3)/(2*2) =1
x =(1-3)/(2*2) = -1/2 ;
2x² -x -1 =2(x+1/2) (x-1).
================================
+ _ +
-------------- -1/2 ---------- 1 ---------------
x∈ (-1/2 ; 1).
================================
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Answers & Comments
Verified answer
1) x^4 -4x³ +4x² -4 ≥ 0;x²(x² -4x +4) -4 ≥ 0;
(x(x-2))² -2² ≥ 0 ;
(x(x-2) -2)(x(x-2) + 2) ≥0;
(x² -2x -2)(x² -2x +2)≥ 0;
т.к. x² -2x+2 =(x-1)² +1 ≥1 >0,
то x² -2x -2 ≥ 0 ; [ (x-x₁)(x-x₂) ≥ 0 ] ***
( x -1 + √5)(x -1 -√5) ≥ 0 ;
x∈( - ∞ ; 1-√5] U [ 1+√5 ; ∞) .
===================================
+ - +
----------------- 1 -√5 ---------- 1+√5 ------------
================================
2) (2x² -x)² <1;
(2x² -x)² -1 <0 ;
(2x² -x -1)(2x² - x +1) <0 ;
т.к. 2x² - x +1 =2(x-1/4)² +7/8 ≥7/8 >0;
[ D =1² -4*2*1 = -7 <0; 2>0 ⇒ 2x² - x +1 >0. ]
то 2x² -x -1 <0;
2(x+1/2)(x-1) <0 ;
x∈ (-1/2 ; 1).
===============================
2x² -x -1 = 0 ;
D =1²-4*2(-1) =9 =3² ;
x =(1+3)/(2*2) =1
x =(1-3)/(2*2) = -1/2 ;
2x² -x -1 =2(x+1/2) (x-1).
================================
+ _ +
-------------- -1/2 ---------- 1 ---------------
x∈ (-1/2 ; 1).
================================