128 - 3 ( x + 4 ) = 23

         3 ( x + 4 ) = 128 - 23

         3 ( x + 4 ) = 105

              x + 4   = 105 : 5

              x + 4   = 35

              x         = 35 - 4

              x         = 31

128 - 3 ( x + 4 ) = 23

         3 ( x + 4 ) = 128 - 23

         3 ( x + 4 ) = 105

              x + 4   = 105 : 5

              x + 4   = 35

              x         = 35 - 4

              x         = 31

k mik nha

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

Học tốt

a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0

x + 1 + x + 2 + ... + x + 199 = 0

( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0 

199x + 19900 = 0

199x = 0 - 19900

199x = -19900

x = -19900 : 199

x = -100

Vậy ...

b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11

x - 30 + x - 29 + x - 28 = 11

( x + x + x ) - ( 30 + 29 + 28 ) = 11

3x -  87 = 11

3x = 11 + 87 

3x = 98

x = \(\frac{98}{3}\)

Vậy ... 

a,

128-3x-12=23

3x=128-12-23

3x=93

x=93:3

= 31

b,

(12x+84+55):5=35

12x+84+55=35.5

12x+84+55=175

12x=175-55-84

12x=36

x=36:12

x=3

\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)

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) \(720:\left[41-\left(2x-5\right)\right]:5=35\)

\(720:\left(41-2x+5\right):5=35\)

\(720:\left(46-2x\right)=35.5\)

\(720:\left(46-2x\right)=175\)

\(46-2x=720:175\)

\(46-2x=\dfrac{144}{35}\)

\(2x=46-\dfrac{144}{35}\)

\(2x=\dfrac{1466}{35}\)

\(x=\dfrac{1466}{35}:2\)

\(x=\dfrac{733}{35}\)

128 - 3( x + 4 ) = 23

3( x + 4 ) = 105

x + 4 = 35

x = 31

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

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