\(\frac{3x-4}{8}\)+\(\frac{6-3x}{8}\)
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NV
1
NH
20 tháng 3 2020
\(\frac{2x-8}{6}-\frac{3x-1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)
\(\Leftrightarrow\frac{2x-8}{6}-\frac{3x-1}{4}-\frac{9x-2}{8}-\frac{3x-1}{12}=0\)
\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x-1\right)}{24}-\frac{3\left(9x-2\right)}{24}-\frac{2\left(3x-1\right)}{24}=0\)
\(\Leftrightarrow\frac{8x-32-18x+6-27x+6-6x+2}{24}=0\)
\(\Leftrightarrow\frac{-43x-18}{24}=0\)
\(\Rightarrow-43x-18=0\)
\(\Leftrightarrow-43x=18\)
\(\Leftrightarrow x=\frac{-18}{43}\)
Vậy...
A
26 tháng 4 2020
\(\frac{2x-8}{6}-\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)
\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x+1\right)}{24}=\frac{3\left(9x-2\right)}{24}+\frac{2\left(3x-1\right)}{24}\)
\(\Leftrightarrow\frac{8x-32}{24}-\frac{18x+6}{24}=\frac{27x-6}{24}+\frac{6x-2}{24}\)
\(\Leftrightarrow8x-32-18x-6=27x-6+6x-2\)
\(\Leftrightarrow8x-18x-27x-6x=-6-2+32+6\)
\(\Leftrightarrow-42x=30\)
\(\Leftrightarrow x=-\frac{5}{7}\)
KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
5 tháng 3 2020
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
IN
4
30 tháng 6 2019
\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}=\frac{3x+5}{8}\)
\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}-\frac{3x+5}{8}=0\)
\(\left(3x-5\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\right)=0\)
\(\Rightarrow x=-\frac{5}{3}\)
AM
21 tháng 3 2020
a.ĐK: 2x2+1\(\ne0\) \(\forall x\)
Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x2+1 >0 với mọi x)
\(\Leftrightarrow x=2\) (TM)
Vậy ...
b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)
Để phương trình bằng 0 thì x2-x-6=0 (Vì x-3\(\ne0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)
Vậy ...
c. ĐK: x\(\ne\)2
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)
\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)
\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)
Vậy ...
d. ĐK: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)
\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))
\(\Leftrightarrow x=-2\:\left(TM\right)\)
Vậy...
9 tháng 3 2019
\(\frac{6}{9}=\frac{8}{3x-4}\)
\(\Rightarrow\frac{2}{3}=\frac{8}{3x-4}\)
\(\Rightarrow\frac{8}{12}=\frac{8}{3x-4}\)
\(\Rightarrow3x-4=12\)
\(\Rightarrow3x=12+4\)
\(\Rightarrow3x=16\)
\(\Rightarrow x=16:3\)
\(\Rightarrow x=\frac{16}{3}=5\frac{1}{3}\)
Vậy \(x=5\frac{1}{3}\)
ta co 3x-4+6-3x=2
->(3x-4+6-3x)/8=2/8=1/4