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A
1
A
0
LN
1
13 tháng 9 2018
a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)
\(\frac{X}{5}=\frac{1}{5}\)
Vậy \(X=1\)
b)\(X-\frac{41}{5}=\frac{-2}{3}\)
\(X=\frac{-2}{3}+\frac{41}{5}\)
\(X=\frac{113}{15}\)
Vậy \(X=\frac{113}{15}\)
c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)
\(\frac{31}{5}-X=\frac{131}{30}\)
\(X=\frac{31}{5}-\frac{131}{30}\)
\(X=\frac{11}{6}\)
Vậy \(X=\frac{11}{6}\)
d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)
\(\frac{9}{X}=\frac{1}{20}\)
\(X=9:\frac{1}{20}\)
\(X=180\)
Vậy \(X=180\)
Hc tốt
TT
1
2 tháng 9 2015
x-2+7 = 1.3.-9
=> x + 5 = -27
=> x = -32
4x - 1 = (-5)
=> 4x = -4
x = -1
-2x + 5 = 7
=> -2x = 2
x= -1
17 tháng 10 2020
a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)
=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)
Vậy x = 2/5
b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)
=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)
Vậy x = -6/11
c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)
Vậy x = -13/4
d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)
=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)
=> \(3\left(229+45x\right)=4\left(61+30x\right)\)
=> \(687+135x=244+120x\)
=> \(687+135x-244-120x=0\)
=> \(\left(687-244\right)+\left(135x-120x\right)=0\)
=> \(443+15x=0\)
=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)
Vậy x = -443/15
AP
4
22 tháng 6 2016
a, x + 1/3 = 3/4
- > x = 3/4 - 1/3 = 5/12
b, x - 2/5 = 5/4
-> x = 5/4 + 2/5 = 33/20
c, -x - 2/3 = 6/7
-> -x = 6/7 + 2/3 = 32/21
-> x = -32/21
d, 4/7 - x = 1/3
x = 4/7 - 1/3 = 5/21
22 tháng 6 2016
a) x+1/3= 3/4
x= 3/4 - 1/3
x= 5/12
b) x-2/5 = 5/4
x= 5/4 + 2/5
x = 33 / 20
c) -x - 2/3 = 6/7
-x = 6/7 + 2/3
-x = 32/21 => x = -32/21
d) 4/7 -x = 1/3
x= 4/7 -1/3
x = 5/21
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)