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a) \(128-3\left(x+4\right)=23\)
\(\Rightarrow3\left(x+4\right)=128-23\)
\(\Rightarrow3\left(x+4\right)=105\)
\(\Rightarrow x+4=35\)
\(\Rightarrow x=35-4\)
\(\Rightarrow x=31\)
b) \(\left[\left(4x+28\right)\cdot3+55\right]:5=35\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=35\cdot5\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=175\)
\(\Rightarrow\left(4x+28\right)\cdot3=120\)
\(\Rightarrow4x+28=40\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=3\)
a, \(128-3\left(x+4\right)=23\)
\(=>3\left(x+4\right)=128-23\)
\(=>3\left(x+4\right)=105\)
\(=>x+4=105:3\)
\(=>x+4=35\)
\(=>x=35-4\)
\(=>x=31\)
b, \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(=>\left(4x+28\right).3+55=35.5\)
\(=>\left(4x+28\right).3+55=175\)
\(=>\left(4x+28\right).3=175-55\)
\(=>\left(4x+28\right).3=120\)
\(=>4x+28=120:3\)
\(=>4x+28=40\)
\(=>4x=40-28\)
\(=>4x=12\)
\(=>x=12:4\)
\(=>x=3\)
\(#WendyDang\)
\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
a: =>3(x+4)=115
=>x+4=115/3
hay x=103/3
b: =>\(8^3\cdot\left(12x-64\right)=4\cdot8^4\)
\(\Leftrightarrow12x-64=32\)
=>12x=96
hay x=8
c: \(\Leftrightarrow\left(4x+28\right)\cdot23+55=175\)
=>(4x+28)x23=120
=>4x+28=120/23
=>4x=-524/23
hay x=-131/23
128 - 3(x + 4) = 23
=> 3(x + 4) = 128 - 23
=> 3(x + 4) = 105
=> x + 4 = 105 : 3
=> x + 4 = 35
=> x = 35 - 4
=> x = 31
Vậy x = 31
[(4x + 28) . 3 + 55] : 5 = 35
=> (4x + 28) . 3 + 55 = 35 . 5
=> (4x + 28) . 3 + 55 = 175
=> (4x + 28) . 3 = 175 - 55
=> (4x + 28) . 3 = 120
=> 4x + 28 = 120 : 3
=> 4x + 28 = 40
=> 4x = 40 - 28
=> 4x = 12
=> x = 12 : 4
=> x = 3
Vậy x = 3
1. 128 - 3.(x+4) = 23
3.(x+4) = 5
x+4 = 5 : 3
x+4 = 5/3
x = -7/3
thiếu đề thì làm sao mà làm được . Lần sau ghi đủ và đúng đề thì mk mới làm được nha
a, 128 – 3(x+4) = 23
b, 12 x - 4 3 . 8 3 = 4 . 8 4
c, [(4x+28).3+55]:5 = 35
d, 720:[41 – (2x – 5)] = 2 3 . 5
128-3.(x+4) =23
3.(x+4) =128-23
3.(x+4) = 105
x+4 =105:3
x+4 =35
x = 35-4
x=31
[(4 : x + 28).3+55]:5=35
(4:x+28).3+55=35.5=175
(4:x+28).3=175-55=120
4:x+28=120:3=40
4:x=40-28=16
x=4:16
=>x=0,25
=> (4:x+28).3+55=35.5
=> (4:x+28).3=175-55
=> 4:x+28=120:3
=> 4:x=40-28
=> 4:x=12
=> x=4:12
=> x=\(\frac{1}{3}\)