Question

решите рациональные неравенства
11-(x+1)^2≥x
(2x-8)^2-4x(2x-8)≥0
x(x+5)-2>4x
(1/3)x^2+3x+6<0
x>(x^2/2)-4x+5(1/2)

Answer 1

11-(x+1)²≥x

11-x²-2x-1≥x

x²+3x-10≤0

x²+3x-10=0 D=49 √D=7

x₁=2 x₂=-5

(x-2)(x+5)≤0

-∞____+____-5___-____2____+____+∞ ⇒

Ответ: x∈[-5;2]

(2x-8)²-4x*(2x-8)≥0

(2x-8)(2x-8-4x)≥0

(2x-8)(-2x-8)≥0

-(2x-8)*(2x+8)≥0 |÷(-1)

4x²-64≤0 |÷4

x²-16≤0

(x-4)(x+4)≤0

-∞____+____-4____-____4____+____+∞ ⇒

Ответ: x∈[-4;4].

x*(x+5)-2>4x

x²+5x-2-4x>0

x²+x-2>0

x²+x-2=0 D=9 √D=3

x₁=1 x₂=-2 ⇒

(x-1)(x+2)>0

-∞____+____-2____-____1____+____+∞ ⇒

Ответ: x∈(-∞-2)U(1;+∞).

(1/3)*x²+3x+6<0 |×3

x²+9x+18<0

x²+9x+18=0 D=9 √D=3

x₁=-3 x₂=-6 ⇒

(x+3)(x+6)<0

-∞_____+_____-6_____-_____-3____+_____+∞ ⇒

Ответ: x∈(-6;-3).

x>(x²/2)-4x+5¹/₂

x>(x²/2)-4x+11/2 |×2

2x>x²-8x+11

x²-10x+11<0

x²-10x+11=0 D=56 √D=√56

x₁=5-√14 x₂=5+√14

-∞____+_____5-√14____-_____5+√14____+____+∞ ⇒

Ответ: x∈(5-√14;5+√14).

Answer 2

$1); ; 11-(x+1)^2geq x\\11-x^2-2x-1geq x\\x^2+3x-10leq 0; ,; ; x_1=-5; ,; x_2=2; (teorema; Vieta)\\znaki:quad +++[-5, ]---[, 2, ]+++\\xin [-5,2, ]\\2); ; (2x-8)^2-4x(2x-8)geq 0\\4x^2-32x+64-8x^2+32xgeq 0\\-4x^2+64geq 0; |:(-4); ; to ; ; x^2-16leq 0\\(x-4)(x+4)leq 0; ,qquad +++[-4, ]---[, 4, ]+++\\xin [-4,4, ]\\3); ; x(x+5)-2>4x\\x^2+5x-4x-2>0; ; to \\x^2+x-2>0; ,; ; x_1=-2; ,; x_2=1; (teorema; Vieta)\\(x+2)(x-1)>0quad +++(-2)---(1)+++\\xin (-infty ,-2)cup (1,+infty )$

$4); ; frac{1}{3}x^2+3x+6<0; |cdot 3\\x^2+9x+18<0; ,; ; x_1=-6; ,; x_2=-3; ; (teorema; Vieta)\\(x+6)(x+3)<0quad ++(-6)---(-3)+++\\xin (-6,-3)\\5); ; x>frac{x^2}{2}-4x+5frac{1}{2}; |cdot 2\\2x>x^2-8x+11; ,\\x^2-10x+11<0; ,; D/4=5^2-11=14; ,; ; x_{1,2}=5pm sqrt{14}\\znaki:quad +++(5-sqrt{14})---(5+sqrt{14})+++\\xin (5-sqrt{14}; ;; 5+sqrt{14})$