Tim x
\(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)
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Tim x biet
\(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)
Tim x biet
c) \(x-3\frac{1}{2}x=-2\frac{6}{7}\)
d) \(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)
tim xa)2frac{3}{4}x-1frac{5}{8}x1b)2frac{3}{4}x-1frac{5}{8}x1c)frac{3}{2}x-frac{2}{5}frac{1}{3}x-frac{1}{4}d)2x-frac{1}{4}frac{5}{6}-frac{1}{2}xe)frac{-5}{6}+3xfrac{2}{3}-frac{1}{2}xf)frac{5}{2}x-frac{3}{2}x+frac{29}{10}g)frac{2}{3}+frac{7}{3}xfrac{5}{4}x+frac{1}{6}h)frac{1}{3}.x+frac{2}{5}left(x-1right)0giup mk nha moi nguoi
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tim x
a)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
b)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
c)\(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
d)\(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\)
e)\(\frac{-5}{6}+3x=\frac{2}{3}-\frac{1}{2}x\)
f)\(\frac{5}{2}x-\frac{3}{2}=x+\frac{29}{10}\)
g)\(\frac{2}{3}+\frac{7}{3}x=\frac{5}{4}x+\frac{1}{6}\)
h)\(\frac{1}{3}.x+\frac{2}{5}\left(x-1\right)=0\)
giup mk nha moi nguoi
Tim so huu ti x
a)\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\))
b)\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
c)\(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{2}{9-6x}-\frac{3}{2}\)
e)\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\)
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
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Tim x:
\(x-\frac{\frac{x}{2}-\frac{x}{3}}{4}-\frac{x}{6}=\frac{2\left(1+x\right)}{3}-\frac{\frac{x}{3}+\frac{1-x^7}{4}}{2}\)
Câu 1:frac{1}{2}+frac{5}{6}.3frac{2}{5}frac{1}{6}.left(-2frac{3}{5}right)+1frac{2}{3}.left(frac{-13}{5}right)left(3-frac{2}{3}+frac{4}{3}right):left(2frac{1}{3}-2,5right)^2sqrt{16.}|5-2.4|.sqrt{0,09}Câu 2: Tim x, y biêt-frac{3}{7}+xfrac{1}{3}|x+frac{1}{7}|-frac{2}{3}0frac{x-2}{x+3}frac{x-3}{x+1}
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Câu 1:
\(\frac{1}{2}+\frac{5}{6}.3\frac{2}{5}\)
\(\frac{1}{6}.\left(-2\frac{3}{5}\right)+1\frac{2}{3}.\left(\frac{-13}{5}\right)\)
\(\left(3-\frac{2}{3}+\frac{4}{3}\right):\left(2\frac{1}{3}-2,5\right)^2\)
\(\sqrt{16.}|5-2.4|.\sqrt{0,09}\)
Câu 2: Tim x, y biêt
\(-\frac{3}{7}+x=\frac{1}{3}\)
\(|x+\frac{1}{7}|-\frac{2}{3}=0\)
\(\frac{x-2}{x+3}=\frac{x-3}{x+1}\)
1 tìm x biết ;a, 0-|x + 1| 5 b, 2 - | frac{3}{4}- x | frac{7}{12}c, 2 | frac{1}{2}x - frac{1}{3}| - frac{3}{2} frac{1}{4}d, | x - frac{1}{3}| frac{5}{6}e, frac{3}{4}- 2 | 2x - frac{2}{3}| 2f, frac{2x-1}{2} frac{5+3x}{3}
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1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
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a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
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1 tìm x biết ;a, 0-|x + 1| 5 b, 2 - | frac{3}{4}- x | frac{7}{12}c, 2 | frac{1}{2}x - frac{1}{3}| - frac{3}{2} frac{1}{4}d, | x - frac{1}{3}| frac{5}{6}e, frac{3}{4}- 2 | 2x - frac{2}{3}| 2f, frac{2x-1}{2} frac{5+3x}{3}
Đọc tiếp
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)