giúp mik câu này với :((
|4:x-\(\frac{1}{2}\)|=3
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NT
0
NM
4
3 tháng 7 2019
Ta có: \(\frac{2-x}{4}=\frac{3x-1}{3}\)
\(\Leftrightarrow\left(2-x\right).3=\left(3x-1\right).4\)
\(\Leftrightarrow6-3x=12x-4\)
\(\Leftrightarrow-3x-12x=-4-6\)
\(\Leftrightarrow-15x=-10\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy ...
NN
3 tháng 7 2019
\(\frac{2-x}{4}=\frac{3x-1}{3}\)
\(3.\left(2-x\right)=4.\left(3x-1\right)\)
\(6-3x=12x-4\)
\(-3x-12x=-4-6\)
\(-15x=-10\)
\(x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
H
2 tháng 5 2023
\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
A
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)
\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
x+2010=0
x=-2010
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
13 tháng 7 2017
f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
12 tháng 4 2018
\(\frac{3}{4}+\frac{1}{4}:X=\frac{1}{2}\)
=> \(\frac{1}{4}:X=\frac{1}{2}-\frac{3}{4}\)
=> \(\frac{1}{4}:X=\frac{4}{8}-\frac{6}{8}\)
=> \(\frac{1}{4}:X=\frac{-1}{4}\)
=>\(X=\frac{1}{4}:\frac{-1}{4}\)
=> X = -1
HT
4
23 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}x=\frac{3}{2}\)
\(\frac{1}{4}x=\frac{3}{2}-\frac{1}{2}\)
\(\frac{1}{4}x=1\)
\(x=1:\frac{1}{4}\)
\(x=4\)
23 tháng 8 2018
\(\frac{1}{2}\)+\(\frac{1}{4}\)x = \(\frac{3}{2}\)
=> \(\frac{1}{4}\)x = \(\frac{3}{2}\)- \(\frac{1}{2}\)
=> \(\frac{1}{4}\)x = 1
=> x = 1 : \(\frac{1}{4}\)
=> x = 4
Vậy x = 4
K
0
26 tháng 8 2018
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0
=> x + 100 = 0 => x = -100
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)
\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)
Vì 1/47 + 1/48 - 1/49 khác 0
Nên x -50 = 0 => x = 50
\(\left|4:x-\frac{1}{2}\right|=3\Leftrightarrow\orbr{\begin{cases}4:x-\frac{1}{2}=3\\4:x-\frac{1}{2}=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}\frac{4}{x}=\frac{7}{2}\\\frac{4}{x}=-\frac{5}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{8}{7}\\x=-\frac{8}{5}\end{cases}}}\)
Vì | 4:x - \(\frac{1}{2}\)| = 3 nên ta có 2 trường hợp
Trường hợp 1:
4:x - \(\frac{1}{2}\)= 3
4: x = 3 + \(\frac{1}{2}\)
4: x = \(\frac{7}{2}\)
x = \(\frac{7}{2}\). 4 = 14
Trường hợp 2:
4:x-\(\frac{1}{2}\)= -3
4:x = (-3) + \(\frac{1}{2}\)
4:x = \(\frac{-5}{2}\)
x = \(\frac{-5}{2}\). 4 = -10
Vậy x= 14 hoặc x = -10
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