Прошу вас. Решите неравенства. По-фотографии.. Created by milashcalennova. algebra-ru

Прошу вас. Решите неравенства. По-фотографии...

milashcalennova @milashcalennova

July 2022 1 5 Report

Прошу вас. Решите неравенства. По-фотографии.

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oganesbagoyan

Verified answer

Решите неравенства
---------------------------------
1) x² -(5/6)*x +1/6 ≥ 0 ⇔x² -(1/3 +1/2)*x +(1/3)*(1/2) ≥ 0 ⇔(x -1/3)(x-1/2)  ≥ 0
+    - +
//////////////////// [1/3] ----------------[ 1/2 ] /////////////////////

ответ : x ∈ (-∞; 1/3] ∪[1/2 ;∞) .

* * * ИЛИ  * * *
x² -(5/6)*x +1/6 ≥ 0⇔ 6x² -5x +1 ≥ 0 ;
трехчлен 6x² -5x +1  разложим на линейные множители традиционно ;    для этого сначала решая уравнение   6x² -5x +1=0 найдем его корни .
D = 5² - 4*6*1 = 25 -24 =1²
x₁ =(5-1) / 2*6 = 4 / 12 =1/3.
x₂ =(5+1) / 12 = 1/2.
---
6x² -5x +1 ≥ 0 ⇔6(x -1/3)(x-1/2)  ≥ 0 ⇔(x -1/3)(x-1/2)  ≥0  ⇒
x ∈ (-∞; 1/3] ∪[1/2 ;∞) .
=======
3) 1/2 - (9/4)*x +x² ≤ 0 ⇔x² -(2 +1/4)x + 2*1/4 ≤ 0   ⇔ (x -1/4)(x-2)  ≤ 0
+      -     +
--------------- [ 1/4] //////////////// [ 2] -------------

ответ : x ∈ [ 1/4 ; 2 ] .
* * * ИЛИ * * *
1/2 - (9/4)*x +x²  ≤ 0 ⇔4x² -9x +2 ≤ 0
4x² -9x +2 = 0
D = 9² - 4*4*2 = 81 -32 =49 =7²
x₁ =(9-7) / 2*4 = 2 /8 =1/4 ;
x₂ =(9+7) / 8 = 16/8 =2.
---
4x² -9x +2 ≤ 0 ⇔ 4(x -1/4)(x-2)  ≤ 0 ⇒ x ∈ [ 1/4 ; 2 ] .
=======
5) (3-x) /(x+2)  ≤ 0

/////////// ( -2) -------- [3] /////////////

твет : x ∈ (- ∞ ;- 2) ∪ [ 3  ;∞) .

* * * ИЛИ * * *
⇔ (x-3) / (x+2)  ≤ 0 ⇔{ x ≠ -2  ; (x+2)* (x - 3) ≥ 0   ;
+   - +
////////////////// [ -2] ----------- [-3] ///////////////////

x ∈ (- ∞ ;- 2) ∪ [ 3  ;∞) .

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