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a)\(\frac{x}{15}=\frac{2}{5}\)=> x=\(\frac{15.2}{5}=6\)
b) \(\frac{3}{x^7}hay\frac{3}{x.7}\)
a,x/9=5/3 = x=15
b,17/x=85/105 = x = 21
c,6/8=15/x = x = 20
d,x/8=2/3 = x = 6
Câu 2:
a: \(=\left(-2\right)^3\cdot\left(-3\right)^3\cdot5^3=30^3\)
b: \(=3^3\cdot\left(-2\right)^3\cdot\left(-7\right)\cdot\left(-7\right)^2\)
\(=\left(3\cdot2\cdot7\right)^3=42^3\)
(-18 - 9) x ( 18+9)
= -18 - 9 x 18+9
= ( -18 + 18 ) x ( 9-9 )
= 0x0
=0
135 x 15 + (-135) x 85
= -135 + 135 x ( 15 + 85 )
= 0 x 100
= 0
(-8) x (-3)3 x (+125)
=(-8) x (-27 ) x 125
= 125 x ( -8 ) x ( -27 )
= -1000 x ( -27)
= 27000
k mk na <3
a) 3/x-7 = 27/135
3/x-7 = 3/15
x - 7 = 15
x = 15 + 7
x = 22
a) \(\frac{3}{x-7}=\frac{27}{135}\)
\(\Rightarrow\)\(\left(x-7\right).27=3.135\)
\(\Rightarrow\)\(\left(x-7\right).27=405\)
\(\Rightarrow\)\(x-7=15\)
\(\Rightarrow\)\(x=22\)
Vậy \(x=22\)
b ) \(71+65.4=\frac{x+140}{x}+260\)
\(71+260=\frac{x+140}{x}+260\)
\(331=\frac{x+140}{x}+260\)
\(331-260=\frac{x+140}{x}\)
\(71=\frac{x+140}{x}\)
\(71=\frac{x}{x}+\frac{140}{x}\)
\(71=1+\frac{140}{x}\)
\(70=\frac{140}{x}\)
\(x=140\div70\)
\(x=20\)
Vậy \(x=20\)
#TQY
a, \(x\) + (\(x\) + 1) + (\(x\) + 3) = 153
\(x\) + \(x\) + 1 + \(x\) + 3 = 153
3\(x\) + 4 = 153
3\(x\) = 153 - 4
3\(x\) = 149
\(x\) = 149 : 3
\(x\) = \(\dfrac{149}{3}\)
b, (\(2x-7\)) - (\(x\) + 135) =0
2\(x\) - 7 - \(x\) - 135 = 0
\(x\) - 142 = 0
\(x\) = 142
x+7=135-(135+89)
=>x+7=135-135-89
=>x+7=0-89=-89
=>x=-89-7
=>x=-96
3/x-7 = 27/135
3/x-7 = 3/15
x - 7 = 15
x = 15 + 7
x = 22