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2x - 1 + 4x - 2 + 6x - 3 + ... + 400x - 200 = 5 + 10 + 15 + ... + 10000
=> (2x + 4x + 6x + ... + 400x) - (1 + 2 + 3 + ... + 200) = 5.(1 + 2 + 3 + ... + 200)
=> 2x.(1 + 2 + 3 + ... + 200) - (1 + 2 + 3 + ... + 200) = 5.(1 + 2 + 3 + ... + 200)
=> (1 + 2 + 3 + ... + 200).(2x - 1) = 5.(1 + 2 + 3 + ... + 200)
=> 2x - 1 = 5
=> 2x = 5 + 1
=> 2x = 6
=> x = 6 : 2
=> x = 3
Vậy x = 3
123 -5 . (x + 4) = 38
5 . (x + 4) = 123 - 38 = 85
x + 4 = 85 : 5 = 17
x = 17 - 4 = 13
(3x - 24) . 73 = 2.74
(3x - 24) = 2.7 = 14
3x - 16 = 14
3x = 14 + 16 = 30
x = 30 : 3 = 10
MẤY DÒNG NÀO BẠN THẤY KO CẦN THIẾT THÌ LƯỢC BỎ NHA!!!
a) \(2\left(x-5\right)-3\left(x+6\right)=4\left(x-7\right)\)
\(2x-10-3x-18=4x-28\)
\(2x-3x-4x-10-18=-28\)
\(-5x-28=-28\)
\(-5x=-28+28=0\)
\(x=\frac{0}{-5}=0\)
b) \(3\left(x-1\right)-2\left(x+5\right)=2\left(x-3\right)\)
\(3x-3-2x-10=2x-6\)
\(3x-2x-2x-3-10=-6\)
\(-x-13=-6\)
\(-x=-6+13=7\)
\(x=-7\)
c) \(5\left(1-x\right)-6\left(1+x\right)=7\left(3-x\right)\)
\(5-5x-6-6x=21-7x\)
\(-5x-6x+7x+5-6=21\)
\(-4x-1=21\)
\(-4x=22\)
\(x=\frac{22}{-4}=\frac{-11}{2}\)
d) \(2x+5-3\left(3x+7\right)=6\left(1-x\right)+8\)
\(2x+5-9x-21=6-6x+8\)
\(2x-9x+6x+5-21=6+8\)
\(-x-16=14\)
\(-x=14+16=30\)
\(x=-30\)
e) \(x-2+3\left(x-4\right)=5\left(x-6\right)+7\)
\(x-2+3x-12=5x-30+7\)
\(x+3x-5x-2-12=-30+7\)
\(-x-14=-23\)
\(-x=-23+14=-9\)
\(x=9\)
f) \(x+2+3\left(1-x\right)-5\left(2-x\right)=6\left(1-x\right)+\left(3-x\right)\)
\(x+2+3-3x-10+5x=6-6x+3-x\)
\(x-3x+5x+6x+x+2+3-10=6+3\)
\(10x-7=9\)
\(10x=9+7=16\)
\(x=\frac{16}{10}=\frac{8}{5}\)
x23:216=x15.x5
x23: 63= x20
==> 63=x23:x20
==> 63=x3
==> x=6
(5x.5x+2):53=125
==> 52x+2:53=53
==> 52x+2=53.53
==> 52x+2=56
==> 2x+2=6
2x=6–2
2x=4
x=4:2
x=2
(3x.3x+4):33=243
==> 32x+4:33=35
32x+4=35.33
32x+4=38
==> 2x+4=8
2x=8–4
2x=4
x=4:2
x=2
3x+3x+1+3x+2=243
3x.1+3x.31+3x.32=243
3x.(1+31+32)=243
3x. (1+3+9)=243
3x.13=243
3x=243:13
3x=...
1/
a)28+2x=35-(-13)
28+2x=48
2x=48-28
x=20:2
x=10
b)2(x+1)2=-7+15
2(x+1)2=8
(x+1)2=4
x+12=22
x+1=2
x=1
c)(-25)-(x+5)=415+5(x-83)
-25-5-x=415+5x-415
-30-x=5x
-30=x+5x=6x
x=-30:6
x=-5
d)|x-2|+(-7)=-3
|x-2|=-3+7
|x-2|=4
=>x-2=4 hoặc x-2=-4
=>x=6 hoặc x=-2
Vậy x=6 hoặc x=-2