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a. 1440 : [ 120 - (3x + 9 ) ] = 120
120 - (3x + 9) = 1440 : 120 = 12
3x + 9 = 120 - 12 = 108
3x = 108 - 9 = 99
x = 99 : 3 = 33
b. 120 + [ ( 999 - 9x ) : 60 ] . 24 = 480
[ ( 999 - 9x ) : 60 ] . 24 = 480 - 120 = 360
( 999 - 9x ) : 60 = 360 : 24 = 15
( 999 - 9x ) = 15 . 60 = 900
9x = 999 - 900 = 99
x = 99 : 9 = 11
390-(x-7)=169:13
390-(x-7)=13
X-7=390-13
X-7=377
X=377+7
X=384
phần a bạn ở dưới làm r nhé
phần b chép sai đề
phần c :
x - 6 : 2 - ( 48 - 24 ) : 2 : 6 - 3 = 0
<=>x - 3 - 24 : 12 - 3 = 0
<=>x - 3 - 2 - 3 = 0
<=>x=8
Vậy x = 8
phần d :
x + 5 . 2 - ( 32 + 16 . 3 : 16 - 15 ) = 0
<=>x + 10 - ( 32 + 3 - 15 ) = 0
<=> x + 10 - 20 = 0
<=>x = 10
Vậy x = 10
hình như bài làm r bạn ạ :0 mình giải r đấy xem lại nhé :V
a) \(390-\left(x-7\right)=169:13\)
\(390-\left(x-7\right)=13\)
\(x-7=390-13\)
\(x-7=377\)
\(x=384\)
b) \(\left(x-110\right):7-3^3-2^3.3\)
Đề sai nhé bạn
c) \(x-6:2-\left(48-24\right):2:6-3=0\)
\(x-3-\left(24\right):2:6=3\)
\(x-3-\left(24\right):2=18\)
\(x-3-\left(24\right)=36\)
\(x-3=36+24\)
\(x-3=60\)
\(x=63\)
d) \(x+5.2-\left(32+16.3:16-15\right)=0\)
\(x+10-\left(32+48:16-15\right)=0\)
\(x+10-\left(32+3-15\right)=0\)
\(x+10-20=0\)
\(x+10=20\)
\(x=10\)
Tìm x
a.( x - 140 ) : 3 = 27
x - 140 = 27 . 3
x - 140 = 81
x = 221
b.14 - 4 ( x + 1 ) = 10
4 ( x + 1 ) = 14 - 10
4 ( x +1) = 4
x + 1 = 1
x = 0
c. 15 ( 7 - x ) = 15
7 - x = 1
x = 6
d.34 ( x - 3 ) = 0
\(\Rightarrow\) 34 = 0 hoặc x - 3 = 0
1. 34 = 0 ( vô lí )
2. x - 3 = 0 \(\Rightarrow\) x = 3
e. 24 + 6 (3 - x ) = 30
6( 3- x ) = 30 - 24
6( 3 - x ) = 6
3 - x = 1
x = 2
f. x3 + 24 = 51
x3 = 51 - 24
x3 = 27
\(\Rightarrow\)x = 3 ; x = -3
g. ( x- 5 )2 - 5 = 44
( x - 5) 2 = 49
\(\Rightarrow\)x - 5 = 7 hoặc x - 5 = -7
1. x - 5 = 7\(\Rightarrow\)x = 12
2. x - 5 = -7 \(\Rightarrow\)x = -2
h. ( x + 1 )3 - 23 = 4
( x + 1 )3 =27
\(\Rightarrow\) x + 1 = 3 hoặc x + 1 = -3
1. x + 1 = 3\(\Rightarrow\)x = 2
2. x + 1 = -3 \(\Rightarrow\)x = -4
a.
\(\left(3x-2^4\right)\times7^3=2\times7^4\)
\(3x-2^4=2\times7^4\div7^3\)
\(3x-16=2\times7\)
\(3x-16=14\)
\(3x=14+16\)
\(3x=30\)
\(x=\frac{30}{3}\)
\(x=10\)
b.
\(2130-\left(x+130\right)+72=-64\)
\(2130-\left(x+130\right)=-64-72\)
\(2130-\left(x+130\right)=-136\)
\(x+130=2130+136\)
\(x+130=2266\)
\(x=2266-130\)
\(x=2136\)
c.
\(15-\left(-x+18\right)=-24\)
\(\left(-x+18\right)=15+24\)
\(-x+18=39\)
\(-x=39-18\)
\(-x=21\)
\(x=-21\)
d.
\(10-\left|x+3\right|=-4-\left(-10\right)\)
\(\left|x+3\right|=10+4-10\)
\(\left|x+3\right|=4\)
\(x+3=\pm4\)
TH1:
\(x+3=4\)
\(x=4-3\)
\(x=1\)
TH2:
\(x+3=-4\)
\(x=-4-3\)
\(x=-7\)
Vậy x = 1 hoặc x = - 7.
a)\(\left(x-140\right)\)\(\div\)7=27-4-3 =20\(\Rightarrow\) \(\left(x-140\right)\) =20x7=140\(\Rightarrow\) x=0
\(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
<=> \(\left(19x+50\right):14=25-16\)
<=> \(\left(19x+50\right):14=9\)
<=> \(19x+50=126\)
<=> \(19x=76\)
<=> \(x=4\)
câu B làm j có \(x\)để tìm
\(390-\left(x-7\right)=169:13\)
<=> \(390-x+7=13\)
<=> \(390-x=6\)
<=> \(x=384\)
\(x-6:2-\left(48:24\right):2:6-3=0\)
<=> \(x-3-2:2:6=3\)
<=> \(x-3-\frac{1}{6}=3\)
<=> \(x=\frac{37}{6}\)
\(x+5.2-\left(32+16.3:6-15\right)=0\)
<=> \(x+10-25=0\)
<=> \(x=15\)
a: 48-3(x+5)=24
=>3(x+5)=48-24=24
=>\(x+5=\dfrac{24}{3}=8\)
=>x=8-5=3
b: \(2^{x+1}-2^x=32\)
=>\(2\cdot2^x-2^x=32\)
=>\(2^x=32=2^5\)
=>x=5
c: \(\left(15+x\right):3=3^3\)
=>\(x+15=3^3\cdot3=3^4=81\)
=>x=81-15=66
d: \(250-10\left(24-3x\right):15=224\)
=>\(\dfrac{2}{3}\left(24-3x\right)=250-224=26\)
=>\(24-3x=26:\dfrac{2}{3}=26\cdot\dfrac{3}{2}=39\)
=>3x=24-39=-15
=>\(x=-\dfrac{15}{3}=-5\)