Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
a) \(\left|48-3x\right|=0\)
\(\Leftrightarrow48-3x=0\)
\(\Leftrightarrow3x=48\)
\(\Leftrightarrow x=\dfrac{48}{3}=16\)
Vậy x = 16.
b) \(\left|-x-7\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24\\-x-7=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=31\\-x=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-31\\x=17\end{matrix}\right.\)
Vậy \(x=-31\) hoặc \(x=17\).
c) \(\left|4-x\right|=21\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21\\4-x=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=25\end{matrix}\right.\)
Vậy \(x=-17\) hoặc \(x=25\).
d) \(\left|x+8\right|+12=0\)
\(\Leftrightarrow\left|x+8\right|=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12\\x+8=-\left(-12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-20\\x=4\end{matrix}\right.\)
Vậy \(x=-20\) hoặc \(x=4\).
e) \(6-\left|x\right|=2\)
\(\Leftrightarrow\left|x\right|=6-2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-4\).
Chúc bạn học tốt!
1. |48 - 3x| = 0.
\(\Leftrightarrow\) 48 - 3x = 0.
\(\Leftrightarrow\) 3x = 48.
\(\Leftrightarrow\) x = \(\dfrac{48}{3}=16.\)
Vậy x = 16.
2. |-x - 7| = 24.
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24.\\-x-7=-24.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=31.\\-x=-17.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
3. |4 - x| = 21.
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21.\\4-x=-21.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
4. |x + 8| + 12 = 0.
|x + 8| = 0 - 12.
|x + 8| = -12.
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=-\left(-12\right).\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=12.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
5. 6 - |x| = 2.
|x| = 6 - 2.
|x| = 4.
\(\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
( 2x + 1 / 2 ) * ( 3 / 4 - x ) = 0
=> 2x + 1 / 2 = 0 hoặc 3 / 4 - x = 0
2x = - 1 / 2 x = 3 / 4
x = - 1 / 4
Vậy x = -1 / 4 hoặc x = 3 / 4
2 / 10 * 12 + 2 / 12 * 14 + 2 / 14 * 16 + ... + 2 / 48 * 50
= 1 / 10 - 1 / 12 + 1 / 12 - 1 / 14 + 1 / 14 + 1 / 16 + .... + 1 / 48 - 1/ 50
= 1 / 10 - 1 / 50
= 2 / 25
Tìm x : \(\left(2x+\frac{1}{2}\right)\left(\frac{3}{4}-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{2}=0\\\frac{3}{4}-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{4}\end{cases}}}\)
Tính nhanh : \(\frac{2}{10.12}+\frac{2}{12.14}+...+\frac{2}{48.50}\)
\(=2\left(\frac{1}{10.12}+\frac{1}{12.14}+...+\frac{1}{48.50}\right)\)
\(=2.\frac{1}{2}\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
b. 1500(x-7)=0
x-7=0
x=7
c. (2x-4)(48-12x)=0
2x-4=0 hoặc 48-12x=0
x=2 hoặc x=4
d. (x+12)(x-1)=0
x+12=0 hoặc x-1=0
x=-12 hoặc x=1
bài 2 :
a . 128-3(x+4)=23
3(x+4)=105
x+4=35
x=31
b. [(14X+26).3+55]:5=35
(14x+26).3+55=175
(14x+26).3=120
14x+26=40
14x=14
x=1
d. 720:[41-(2X-5)]=23.5
41-(2x-5)=720:(23.5)
41-(2x-5)=144/23
2x-5=799/23
2x=914/23
x=457/23
b, 1500.(x – 7) = 0
<=>1500x-10500=0
<=>1500x=10500
<=>x=7
Vậy x=7
c,(2.x – 4).(48 – 12.x) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy x=2 hoặc x=4
d, (x + 12).(x – 1) =0
\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)
Vậy x=-12 hoặc x=1
Bài 2:
a) 128- 3(x+ 4) = 23
\(\Leftrightarrow\)128-(3x+12)=23
\(\Leftrightarrow\)128-3x-12=23
\(\Leftrightarrow\)116-3x=23
\(\Leftrightarrow\)3x=116-23
\(\Leftrightarrow\)3x=93
\(\Leftrightarrow\)x=31
Vậy x=31
b) [(14x+ 26). 3+ 55]: 5= 35
\(\Leftrightarrow\)(14x+ 26). 3+ 55=175
\(\Leftrightarrow\)42x+78+55=175
\(\Leftrightarrow\)42x+133=175
\(\Leftrightarrow\)42x=175-133
\(\Leftrightarrow\)42x=42
\(\Leftrightarrow\)x=1
Vậy x=1
d, 720: [41- (2x- 5)]= 23. 5
\(\Leftrightarrow\)720: 41- (2x- 5)=115
\(\Leftrightarrow\)41-(2x- 5)=720:115
\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)
\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)
\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)
\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)
Vậy x=\(\dfrac{457}{23}\)
1/ -7264 + (1543 + 7264)
=-7264 + 1543 + 7264=1543
2/ (144 – 97) – 144
=144-97-144=-97
3/ (-145) – (18 – 145)(Vì có dấu trừ ở trước ngoặc nên p đổi dấu)
=-145-18+145=-18
4/ 111 + (-11 + 27)
=111-11+27=137
7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0
7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)
7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)
7/48 - 193/56 : x = 0
193/56 : x = 0 + 7/48
193/56 : x = 7/48
x = 193/56 : 7/48
x = 1158/49
1) x - 43 = (35 - x) - 48
=> x + x = 35 - 48 + 43
=> x + x = 30
=> x = 30 : 2
=> x = 15
2) 305 - x + 14 = 48 + (x + 23)
=> 305 - x + 14 = 48 + x + 23
=> -x - x = 48 + 23 - 14 - 305
=> -x - x = -248
=> -x = -248 : 2
=> -x = -124
=> x = 124
3) - (x - 6 + 85) = (x + 51) - 54
=> -x + 6 - 85 = x + 51 - 54
=> -x - x = 51 - 54 + 85 - 6
=> -x - x = 76
=> -x = 76 : 2
=> -x = 38
=> x = -38
4) - (35 - x - 37 - x) = 33 - x
=> -35 + x + 37 + x = 33 - x
=> x + x + x = 33 + 35 - 37
=> x + x + x = 31
=> x = 31 : 3
=> x \(=\dfrac{31}{3}\)
Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.
5) 13 - | x | = | -4 |
=> 13 - |x| = 4
=> |x| = 13 - 4
=> |x| = 9
=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
6) | x | - 3 + 6 = 16
=> |x| = 16 - 6 + 3
=> |x| = 13
=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)
7) 35 - | 2x - 1 | = 14
=> |2x - 1| = 35 - 14
=> |2x - 1| = 21
=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)
8) | 3x - 2 | + 5 = 9 - x
=> |3x - 2| = 9 - 5 - x
=> |3x - 2| = 4 - x
=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)
Vì x \(\in\) Z nên x = -1.
9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )
=> x - (-18) > 12 - 1
=> x + 18 > 11
=> x > 11 - 18
=> x > -7
10) | 17 + ( x - 15 ) | < 4
=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)
11) x2 - 5x = 0
=> x . (2 - 5) = 0
=> x . (-3) = 0
=> x = 0 : (-3)
=> x = 0
12) | x-9 | . (-8) = -16
=> |x - 9| = (-16) : (-8)
=> |x - 9| = 3
=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)
13) | 4 - 5x | = 24 với x < hoặc = 0
=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vì x \(\le\) 0 nên x = -4
14) x . ( x - 2 ) > 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x>2\\x< 2\end{matrix}\right.\)
15) x . ( x - 2 ) < 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)
16) (x-1) . (y+1) = 5
=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
17) x . ( y +2 ) = -8
=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=2\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-2\\y+2=4\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=4\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-4\\y+2=2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=2-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
18) xy - 2x - 2y = 0
=> x . (y - 2) - 2y = 0
=> x . (y - 2) - 2y - 4 = -4
=> x . (y - 2) - 2 . (y - 2) = -4
=> (y - 2) . (x - 2) = -4
=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=2\\x-2=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2+2\\x=-2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=4\\x=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-2\\x-2=2\end{matrix}\right.=>\left[{}\begin{matrix}y=-2+2\\x=2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)
19) 2x - 5 \(⋮\) x - 1
=> (2x - 2) - (5 - 2) \(⋮\) x - 1
=> 2(x - 1) - 3 \(⋮\) x - 1
Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1
=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}
=> x \(\in\) {-2; 0; 2; 4}
P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~
b,/x-27/=24
TH1 : x - 27 = 24 => x = 51
TH2 : x - 27 = -24 => x = 3
c,/x+8/+12=0
/x+8/=-12
Vì x ko chứa số âm => \(x\in\varnothing\)
d,3/4-x/=21
/4-x/=7
TH1 : 4 - x = 7 => x = -3
TH2: 4 - x = -7 => x = 11
KL :...
\(\left(2x-4\right).\left(48-12x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(2x-4\right)=0\\\left(48-12x\right)=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=4\\12x=48\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=4\end{array}\right.\)
Vậy x = 2 ; 4
\(\left(2x-4\right)\left(48-12x\right)=0\)
\(\Leftrightarrow24\left(x-2\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=4\end{array}\right.\)
câu này nữa nhé:
(x+12)2.(x-1)=0
(2.x-4).(48-12.x)=0
=>(2.x-4)=0 hoặc (48-12.x)=0
=>2.x-4=0
=>2.x=0+4=4
=>x=4:2=2 ->(1)
48-12.x=0
=>12.x=48-0=48
=>x=48:12
=>x=4 ->(2)
Từ (1);(2)=> x có hai giá trị thỏa mãn: x=2 hoặc x=4
a) ( 2x - 4 ) . ( 48 - 12x ) = 0
\(\Leftrightarrow\begin{cases}2x-4=0\\48-12x=0\end{cases}\)
\(\Rightarrow\begin{cases}2x=0+4\\12x=48-0\end{cases}\)
\(\Rightarrow\begin{cases}2x=4\\12x=48\end{cases}\)
\(\Rightarrow\begin{cases}x=4:2\\x=48:12\end{cases}\)
\(\Rightarrow\begin{cases}x=2\\x=4\end{cases}\)
Vậy x = 2 hoặc x = 4
b) ( x + 12 )2 . ( x - 1 ) = 0
\(\Leftrightarrow\begin{cases}\left(x+12\right)^2=0\\x-1=0\end{cases}\)
\(\Rightarrow\begin{cases}x+12=0\\x=0+1\end{cases}\)
\(\Rightarrow\begin{cases}x=0-12\\x=1\end{cases}\)
\(\Rightarrow\begin{cases}x=-12\\x=1\end{cases}\)
Vậy x = -12 hoặc x = 1