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ГулькаПомогулька
@ГулькаПомогулька
July 2022
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Решите пожалуйста, очень нужно!!!
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oganesbagoyan
Verified answer
1)
а)
(x+4)(x-2)(x-5) >0 ;
методом интервалов :
- + - +
-------- -4 ////////// 2 --------5 ////////////
x∈( - 4 ; 2) U (5 ; ∞) .
--------------------
б) x(x+8)/(x+2)(x-7)
≤0 ;
+ - + - +
---------- [-8 ] ---------( -2) --------- [0] ----------( 7 ) --------
x∈[ - 8 ; -2) U[0 ; ∞7) .
--------------------
в)
1/(x -4) - 4/(x² -4x) >0 ;
(x -4)/x(x-4) >0 ;
{x ≠4 ; 1/x > 0. ⇒ ------0 ////////////// 4 ///////////////
x∈( 0 ; 4) U (4 ; ∞) .
-------------------
г)
4(x-2)²(x+3/4)(x-7) ≥ 0;
+ - - +
/////////// [-3/4] ----------- [2] ---------- [7] ///////////////
x∈( -
∞
; -3/4) U {2} U [ 7 ; ∞) .
-------------------
2)
5/(x² -5x+4) > 8/(x² -5x+4) ;
8/(x² -5x+4) - 5/(x² -5x+4) <0 ;
3/(x² -5x+4) <0 ;
x² -5x+4 <0 ;
(x-1)(x-4) <0 ;
+ - +
--------- 1------------ 4 -----------
x∈( 1;
4) .
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Answers & Comments
Verified answer
1)а) (x+4)(x-2)(x-5) >0 ;
методом интервалов :
- + - +
-------- -4 ////////// 2 --------5 ////////////
x∈( - 4 ; 2) U (5 ; ∞) .
--------------------
б) x(x+8)/(x+2)(x-7) ≤0 ;
+ - + - +
---------- [-8 ] ---------( -2) --------- [0] ----------( 7 ) --------
x∈[ - 8 ; -2) U[0 ; ∞7) .
--------------------
в) 1/(x -4) - 4/(x² -4x) >0 ;
(x -4)/x(x-4) >0 ;
{x ≠4 ; 1/x > 0. ⇒ ------0 ////////////// 4 ///////////////
x∈( 0 ; 4) U (4 ; ∞) .
-------------------
г) 4(x-2)²(x+3/4)(x-7) ≥ 0;
+ - - +
/////////// [-3/4] ----------- [2] ---------- [7] ///////////////
x∈( -∞ ; -3/4) U {2} U [ 7 ; ∞) .
-------------------
2) 5/(x² -5x+4) > 8/(x² -5x+4) ;
8/(x² -5x+4) - 5/(x² -5x+4) <0 ;
3/(x² -5x+4) <0 ;
x² -5x+4 <0 ;
(x-1)(x-4) <0 ;
+ - +
--------- 1------------ 4 -----------
x∈( 1;4) .