(x – 2).(x + 4) = 0
(x -2).( x + 15) = 0
(7-x).( x + 19) = 0
-5 < x < 1
|x| < 3
(x - 3)(x - 5) < 0
(x - 5)(x - 7) < 0
-6x - ( -7) = 25
46 - ( x - 11 ) = - 48
| x – 12| - 11 = 9
56 – ( x – 13 ) = - 58
| x + 19| + 24 = 23
(x – 11).( x + 5) = 0
23 (67 x) = 34
|x +1| – 5 = 10
– 7 + 2x = – 37 – (– 26)
2x ² - 3 = 29
Bài 2
: Cho biểu thức: A = (-m + n – p) – (- m – n – p)
a) Rút gọn A b) Tính giá trị của A khi m = 1; n = -1; p = -2
Làm giúp mik vs ạ
(x – 2).(x + 4) = 0 (x -2).( x + 15) = 0 (7-x).( x + 19) = 0 -5 < x < 1 |x| < 3 (x - 3)(x - 5) < 0 (x - 5)(x - 7) < 0 -6x - ( -7)
By Ivy
`1)(x – 2).(x + 4) = 0`
⇒ \(\left[ \begin{array}{l}x-2=0\\x+4=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
vậy `x=2` hoặc `x=-4` là nghiệm của pt
`2) (x -2).( x + 15) = 0`
\(\left[ \begin{array}{l}x-2=0\\x+15=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2\\x=-15\end{array} \right.\)
`3) (7-x).( x + 19) = 0`
\(\left[ \begin{array}{l}7-x=0\\x+19=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=7\\x=-19\end{array} \right.\)
`4) -5 < x < 1`
`⇒ x ∈ { -4; -3; -2; -1; 0}`
`5) |x| < 3`
`=>`\(\left[ \begin{array}{l}-x<3\\x<-1\end{array} \right.\)
`6) (x – 3)(x – 5) < 0`
`=>th1` ` x- 3 > 0` hay `x – 5 <0`
`⇒ x > 3` hay ` x < 5 ⇒ x = 4`
`⇒th2 x – 3 <0` hay `x – 5 > 0`
`⇒ x < 3` hay `x > 5 ⇒ x = ∅`
Vậy `x = 4`
`7) (x – 5)(x – 7) < 0`
`⇒ TH 1: x- 5 > 0` hay `x – 7 <0`
`⇒ x > 5 hay x < 7 ⇒ x = 6`
`⇒th2: x – 5 <0` hay `x – 7 > 0`
`⇒ x < 5` hay `x > 7 ⇒ x = ∅`
`=> x = 6`
`8) -6x – ( -7) = 25`
`=> – 6x + 7 = 25`
`=>6x = 7 – 25`
`=> x = -18/6`
`⇒ x = -3`
`9) 46 – ( x – 11 ) = – 48`
`⇔ x – 11 = 46 + 48`
`⇔ x = 94 + 11`
`⇒ x = 105`
`10) | x – 12| – 11 = 9`
`=>| x – 12| = 9+ 11`
`=> | x – 12| = 20`
`=>`\(\left[ \begin{array}{l}x-12=20\\x-12=-20\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=32\\x=-8\end{array} \right.\)
`11) 56 – ( x – 13 ) = – 58`
`⇔ x – 13 = 56 + 58`
`⇒ x = 127`
`12)| x + 19| + 24 = 23`
`=>| x + 19| = -1`
`=>`\(\left[ \begin{array}{l}x+19=-1\\x+19=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-20\\x=-18\end{array} \right.\)
`13) (x – 11).( x + 5) = 0`
`=>`\(\left[ \begin{array}{l}x-11=0\\x+5=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=11\\x=-5\end{array} \right.\)
`14) 23(67 x)= 34 `
`=>1541x=34`
`=>x=34/1541`
`15) |x +1| – 5 = 10`
`=>|x-1|=15`
`=>`\(\left[ \begin{array}{l}x+1=15\\x+1=-15\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=14\\x=-16\end{array} \right.\)
`16) – 7 + 2x = – 37 – (– 26)`
`⇔ – 7 + 2x = -37 + 26`
`⇔ 2x = -11 + 7`
`⇒ x = – 4/2`
`⇒ x = -2`
`17) 2x ² – 3 = 29`
`⇔ 2x ² = 32`
`⇔ x ² = 32 /2`
`⇔ x² = 16`
`⇔ x=±4`
`Bài 2: ` `a) A = (-m + n – p) – (- m – n – p)`
`A = – m + n – p + m + n + p = (-m + m) + (n + n) + (-p + p)`
`A = 2n`
`b)` Thay `m = 1; n = -1; p = -2` vào `A` ta có:
`A = (-m + n – p) – (- m – n – p) = [-1 + (-1) – (-2)] – [-1 – (-1) – (-2)] `
`A = -1 -1 + 2 – (- 1 +1 +2) = – 2 + 2 – 2 = – 2`
`Vậy` `A = -2`
Đáp án:
Giải thích các bước giải:
Bài 1:
1/. (x – 2).(x + 4) = 0
⇒ x- 2 = 0 hay x+ 4 = 0
⇒ x = 2 hay x = – 4
2/. (x -2).( x + 15) = 0
⇒x – 2 = 0 hay x + 15 = 0
⇒ x = 2 hay x = -15
3/. (7-x).( x + 19) = 0
⇒ 7 – x = 0 hay x + 19 = 0
⇒ x =7 hay x = -19
4/. -5 < x < 1
⇒ x ∈ { -4; -3; -2; -1; 0}
5/. |x| < 3
⇒ x = 3 hay x = -3
6/. (x – 3)(x – 5) < 0
⇒ trường hợp 1: x- 3 > 0 hay x – 5 <0
⇒ x > 3 hay x < 5 ⇒ x = 4
⇒trường hợp 2: x – 3 <0 hay x – 5 > 0
⇒ x < 3 hay x > 5 ⇒ x = ∅
Vậy x = 4
7/. (x – 5)(x – 7) < 0
⇒ trường hợp 1: x- 5 > 0 hay x – 7 <0
⇒ x > 5 hay x < 7 ⇒ x = 6
⇒trường hợp 2: x – 5 <0 hay x – 7 > 0
⇒ x < 5 hay x > 7 ⇒ x = ∅
Vậy x = 6
8/. -6x – ( -7) = 25
⇔ – 6x + 7 = 25
⇔ 6x = 7 – 25
⇒ x = -18 : 6
⇒ x = -3
9/. 46 – ( x – 11 ) = – 48
⇔ x – 11 = 46 + 48
⇔ x = 94 + 11
⇒ x = 105
10/. | x – 12| – 11 = 9
⇔ | x – 12| = 9+ 11
⇔ | x – 12| = 20
⇒ x – 12 = 10 hay x – 12 = -10
⇒ x = 22 hay x = 2
11/. 56 – ( x – 13 ) = – 58
⇔ x – 13 = 56 + 58
⇔ x = 114 + 13
⇒ x = 127
12/. | x + 19| + 24 = 23
⇔| x + 19| = 23 – 24
⇔| x + 19| = -1
⇒ x + 19 = 1 hay x + 19 = -1
⇒ x = -18 hay x = – 20
13/. (x – 11).( x + 5) = 0
⇒ x – 11 = 0 hay x + 5 = 0
⇒ x = 11 hay x = -5
14/. 23 (67 x) = 34 – Đề bị thiếu dữ iệu phần in đậm
15/. |x +1| – 5 = 10
⇔ |x +1| = 10 + 5
⇔ |x +1| = 15
⇒ x + 1 = 15 hay x + 1 = – 15
⇒ x = 14 hay x = – 16
16/. – 7 + 2x = – 37 – (– 26)
⇔ – 7 + 2x = -37 + 26
⇔ 2x = -11 + 7
⇒ x = – 4 : 2
⇒ x = -2
17/. 2x ² – 3 = 29
⇔ 2x ² = 29 + 3
⇔ 2x ² = 32
⇔ x ² = 32 : 2
⇔ x² = 16
⇔ x² =4²
⇒ x = 4 hay x = – 4
Bài 2: a/. A = (-m + n – p) – (- m – n – p)
A = – m + n – p + m + n + p = (-m + m) + (n + n) + (-p + p)
A = 2n
b/. Thay m = 1; n = -1; p = -2 vào A ta có:
A = (-m + n – p) – (- m – n – p) = [-1 + (-1) – (-2)] – [-1 – (-1) – (-2)]
A = -1 -1 + 2 – (- 1 +1 +2) = – 2 + 2 – 2 = – 2
Vậy A = -2