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a,(x+1)-(x+2)-(x+3)=24
=>x+1-x-2-x-3 =24
=>(x-x-x)+(1-2-3) =24
=> -x-4 =24
=> -x =24+4
=> -x =28
=> x =-28
Vậy x=-28
b,4x+2-3(x-1)=3x-5
=>4x+2-3x+3=3x-5
=>3x-4x+3x =2+3+5
=>2x =10
=>x =5
Vậy x=5
c,x-1-2(x-2)=x-11
=>x-1-2x+4=x-11
=>x-2x-x =-11+1-4
=>-2x =-14
=>x =7
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
GIÚP MÌNH NHANH NHA AI NHANH NHẤT MÌNH SẼ K, MÌNH CẦN GẤP LẮM
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1.
a, (x+50)*2=220
=> 2x+100=220
=> 2x = 120
=> x= 60
b, (2x-75)*12=144
=> 24x-900=144
=> 24x=1044
=> x= 43,5
c, (47-3x)=5
=> 47-3x=5
=> 3x=42
=> x= 14
A. ( x + 50 ) x 2= 220
=> 2x + 100 = 220
=> 2x = 220 - 100
=> 2x = 120
=> x = 120 : 2
=> x = 60
B. ( 2x - 75 ) x 12 = 144
=> (2x - 75) x 12 = 144
=> 2x - 75 = 144 : 12
=> 2x - 75 = 12
=> 2x = 75 + 12
=> 2x = 87
=> x = 87 : 2 = 43,5
C. 47 - 3x = 5
=> 3x = 47 - 5 = 42
=> x = 42 : 3
=> x = 14.
2)1 + 2 + 3 + 4 + 5 +..+ X = 2550
=> \(\frac{x.\left(x+1\right)}{2}\)=2550
=> x.(x+1) = 2550 x 2 = 5100. Mà không có 2 số liên tiếp nào nhân với nhau bằng 5100 nên x không thỏa mãn đề bài.
3)Đề sai nha bạn.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
\(2\left(x-2\right)=x\left(x-2\right)\)
\(\Rightarrow2\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2-x=0\end{cases}\Rightarrow x=2}\)
a)2
b)\(\frac{-11}{4}\)