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a) \(\frac{7}{12}+\frac{x}{15}=\frac{1}{20}\)
=> \(\frac{35}{60}+\frac{4x}{60}=\frac{3}{60}\)
=> 35 + 4x = 3
=> 4x = -32
=> x = -8
b) \(\frac{-7}{x}+\frac{8}{15}=\frac{-1}{20}\)(ĐK \(x\ne0\))
=> \(\frac{-7}{x}=\frac{-1}{20}-\frac{8}{15}\)
=> \(\frac{-7}{x}=\frac{-3}{60}-\frac{32}{60}\)
=> \(\frac{-7}{x}=\frac{-35}{60}\)
=> \(\frac{-7}{x}=\frac{-7}{12}\)
=> x = 12(TM)
a) x - (-7) = -8 + (-2)
x + 7 = -8 - 2
x + 7 = -10
x = -10 - 7
x = -17
b) x + (12 - 15) = /-3/ - 7
x + (-3) = 3 - 7
x - 3 = -4
x = -4 + 3
x = -1
c) (12 - x) - (-10) = -25
(12 - x) + 10 = -25
12 - x = -25 - 10
12 - x = -35
x = 12 - (-35)
x = 12 + 35 = 47
d) (-20) - (15 - x) = -50
15 - x = -20 - (-50)
15 - x = -20 + 50
15 - x = -30
x = 15 - (-30)
x = 15 + 30
x = 45
b, x-96=(443-x)-15
x-96=443-x-15
x-96=428-x
x+x=428+96
2x=524
x=262
a, (x-12)-15=(20-7)-(18+x)
x-12-15=13-18-x
x-27=-5-x
x+x=-5+27
2x=22
x=11
`(x+15)/20+(x-7)/4=(x+12)/2+x/5`
`<=>x+15+5(x-7)=10(x+12)+4x`
`<=>x+15+5x-35=10x+120+4x`
`<=>6x-20=14x+120`
`<=>8x=-140`
`<=>-35/2`
Vậy `S={-35/2}`
\(\dfrac{x+15}{20}+\dfrac{x-7}{4}=\dfrac{x+12}{2}+\dfrac{x}{5}\)
\(\Rightarrow\dfrac{x+15}{20}+\dfrac{5\left(x-7\right)}{20}-\dfrac{10\left(x+12\right)}{20}-\dfrac{4x}{20}=0\)
=> x + 15 + 5x - 35 - 10x - 120 - 4x = 0
=> -8x - 140 = 0
=> -8x = 140
=> x = \(-\dfrac{35}{2}\)
tìm x,biết
5-x-16=40+x
4x-10=15-x
15-x=4x-5
x-15=6+4x
-12+x=5x-20
7x-4=20+3x
5x-7=-21-2x
x+15=20-4x
17-x=7-6x
(x - 12) - 15 = (20 - 7) - (18 + x)
<=> x - 12 - 15 = (20 - 7) - (18 + x)
<=> x - 27 = 20 - 7 - (18 + x)
<=> x - 27 = -x - 5
<=> x = -x - 5 + 27
<=> x = -x + 22
<=> 2x = 22
=> x = 11
=> ( x - 12 - 15 ) - ( 20 - 7 - 18 - x ) = 0
x - 12 - 15 - 20 + 7 + 18 + x = 0
x - 12 - 15 - 20 + 7 + 18 + x = 0
2x - 12 - 15 - 20 + 7 + 18 = 0
2x = 0 - 18 - 7 +20 + 15 + 12
2x = 22
x = 22 : 2
x = 11