1)3-(x+\(\dfrac{5}{7}\))=\(\dfrac{9}{21}\)
2)\(\dfrac{x}{2}\)+\(\dfrac{x}{5}\)=\(\dfrac{17}{10}\)
3)\(\dfrac{1}{2}\)x+\(\dfrac{1}{3}\)-1=3\(\dfrac{1}{3}\)
a)\(\dfrac{2}{3}\)x-1=\(\dfrac{3}{2}\)
b)| 5x - \(\dfrac{1}{2}\)| - \(\dfrac{2}{7}\)= 25%
c)\(\dfrac{x-3}{4}\)=\(\dfrac{16}{x-3}\)
d)\(\dfrac{-8}{13}\)+\(\dfrac{7}{17}+\dfrac{21}{31}\)<x≤\(\dfrac{-9}{14}\)+4+\(\dfrac{5}{-14}\)(xϵZ)
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
Tìm x, biết:
a) \(\dfrac{3}{7}\)x - \(\dfrac{2}{3}\)x = \(\dfrac{10}{21}\)
b) \(\dfrac{7}{35}\) : (x - \(\dfrac{1}{3}\)) = \(-\dfrac{2}{25}\)
c) 3.(x - \(\dfrac{1}{2}\)) - 5. (x + \(\dfrac{3}{5}\)) = -x + \(\dfrac{1}{5}\)
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
a,73x−32x=2110⇒x(73−32)=2110⇒x.−215=2110⇒x=−2b,357:(x−31)=−252⇒51:(x−31)=−252⇒x−31=−25⇒x=−613c,3.(x−21)−5.(x+53)=−x+51⇒3x−23−5x+5=−x+51
\(\dfrac{x-1}{21}=\dfrac{3}{x+1}\)
\(2\dfrac{1}{2}x+x=2\dfrac{1}{17}\)
\(\left(x+\dfrac{1}{4}-\dfrac{2}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(2\dfrac{1}{3}x-1\dfrac{3}{4}x+2\dfrac{2}{3}=3\dfrac{3}{5}\)
Giúp mình với ! Mình cần gấp
a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)
( x-1)(x+1) = 21.3
x2 + x - x -1 = 63
x2 = 63 + 1
x2 = 64
x = + - 8
b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)
x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)
x = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)
x = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)
x = \(\dfrac{10}{17}\)
c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)): \(\dfrac{23}{12}\) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)) = \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)
x - \(\dfrac{5}{12}\) = \(\dfrac{7}{12}\)
x = \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)
x = 1
d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\) = 3\(\dfrac{3}{5}\)
x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\) = \(\dfrac{18}{5}\)
x\(\dfrac{7}{12}\) = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)
x\(\dfrac{7}{12}\) = \(\dfrac{14}{15}\)
x = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)
x = \(\dfrac{8}{5}\)
Tìm x :
a, \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) +x) = \(\dfrac{2}{3}\) b, x+\(\dfrac{2}{3}\) =\(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\) c, x-[\(\dfrac{17}{2}\) -(\(\dfrac{-3}{7}\) + \(\dfrac{5}{3}\) )]=\(\dfrac{-1}{3}\)
d,\(\dfrac{9}{2}\) - [\(\dfrac{2}{3}\) - (x+\(\dfrac{7}{4}\) )] =\(\dfrac{-5}{4}\)
Mọi người giúp mình nha
b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)
\(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)
\(x\) = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{1}{10}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x\) = \(\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x\) = \(\dfrac{1}{4}\)
x = \(\dfrac{2}{3}-\dfrac{1}{4}\)
\(x=\dfrac{5}{12}\)
d, \(\dfrac{9}{2}\) - ( \(\dfrac{2}{3}\) - (\(x\) + \(\dfrac{7}{4}\))] = \(\dfrac{-5}{4}\)
\(\dfrac{9}{2}\) - \(\dfrac{2}{3}\) + \(x\) + \(\dfrac{7}{4}\) = \(-\dfrac{5}{4}\)
\(\dfrac{23}{6}\) + \(x\) = - \(\dfrac{5}{4}\) - \(\dfrac{7}{4}\)
\(x\) = - 3 - \(\dfrac{23}{6}\)
\(x\) = - \(\dfrac{41}{6}\)
Tìm x, biết:
a) \(\dfrac{1}{20}\) - (x - \(\dfrac{8}{5}\)) = \(\dfrac{1}{10}\)
b) \(\dfrac{7}{4}\) - (x + \(\dfrac{5}{3}\)) = \(\dfrac{-12}{5}\)
c) x - [\(\dfrac{17}{2}\) - \(\left(\dfrac{-3}{7}+\dfrac{5}{3}\right)\)] = \(\dfrac{-1}{3}\)
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
\(\dfrac{5}{7}-x=\dfrac{9}{21}\)
\(-x-\dfrac{1}{3}=\dfrac{2}{6}\)
\(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{5}{7}-x=\dfrac{9}{21}\\ \Rightarrow\dfrac{5}{7}-x=\dfrac{3}{7}\\ \Rightarrow x=\dfrac{5}{7}-\dfrac{3}{7}\\ \Rightarrow x=\dfrac{2}{7}\\ b,-x-\dfrac{1}{3}=\dfrac{2}{6}\\ \Rightarrow-x=\dfrac{2}{6}+\dfrac{1}{3}\\ \Rightarrow-x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\\ \dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\ \Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{13}{12}\)
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a) x + \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)
b) x - \(\dfrac{2}{5}\) = \(\dfrac{2}{7}\)
c) \(\dfrac{3}{5}\) - x = \(\dfrac{1}{10}\)
d) x . \(\dfrac{3}{4}\) = \(\dfrac{9}{20}\)
e) x : \(\dfrac{1}{7}\) = 14
f) ( \(\dfrac{1}{4}\) + x ) . \(\dfrac{1}{2}\) = \(\dfrac{2}{5}\)
g) x . \(\dfrac{2}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{9}{12}\)
h) \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
k) \(3\dfrac{4}{5}\) - x = \(\dfrac{18}{5}\)
l) x . \(2\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
m) x . \(\dfrac{6}{11}\) + x . \(\dfrac{5}{11}\) = 2025
n) x . \(\dfrac{14}{9}\) - x . \(\dfrac{7}{9}\) + x . \(\dfrac{5}{9}\) = 2
Các bạn làm theo cách bình thường ở lớp 5 cho mính nhé!
Chú ý: dấu "." là dấu nhân.
a: x+2/5=1/2
=>x=1/2-2/5=5/10-4/10=1/10
b; x-2/5=2/7
=>x=2/7+2/5=10/35+14/35=24/35
c: 3/5-x=1/10
=>x=3/5-1/10=6/10-1/10=5/10=1/2
d: x*3/4=9/20
=>x=9/20:3/4=9/20*4/3=36/60=3/5
e: x:1/7=14
=>x=14*1/7=2
f: =>x+1/4=2/5:1/2=4/5
=>x=4/5-1/4=16/20-5/20=11/20
g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12
=>x=17/12:2/3=17/12*3/2=51/24=17/8
Tìm \(x\) biết:
\(a.x=\dfrac{1}{5}+\dfrac{-3}{7}\) \(b.\dfrac{3}{5}-\dfrac{4}{7}\div x=\dfrac{-9}{10}\) \(c.x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\) \(d.\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)