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14 tháng 4 2023
a: x=-5/11+2/11=-3/11
b: =>x=-3/24+20/24+1/24=18/24=3/4
c: =>5/8-x=1/9+5/4=4/36+45/36=49/36
=>x=5/8-49/36=-53/72
d: =>2/3-x=1/3
=>x=1/3
e: =>1/5:x=12/35
=>x=7/12
18 tháng 7 2017
a) -3x+4+5x=-10-x
-3x+4+5x+10+x=0
(-3x+5x+x)+10=0
3x+10=0
3x=-10
x=\(\dfrac{-10}{3}\)
Vậy x=\(\dfrac{-10}{3}\)
b)-x+1=-3x-8
-x+1+3x+8=0
(-x+3x)+(1+8)=0
2x+9=0
2x=-9
x=\(\dfrac{-9}{2}\)
Vậy x=\(\dfrac{-9}{2}\)
c)8-(x-1)=10+(x+5)
8-x+1=10+x+5
9-x=15+x
9-x-15-x=0
(9-15)-(x+x)=0
-6-2x=0
2x=-6
x=-3
Vậy x=-3
d)100+(x+7)-(-2x+3)=8+(x+100)
100+x+7+2x-3=8+x+100
(x+2x)+(100+7-3)=(8+100)+x
3x+104=108+x
3x+104-108-x=0
(3x-x)+(104-108)=0
2x-4=0
2x=4
x=2
Vậy x=2
DH
18 tháng 7 2017
e, \(\left|2x+5\right|=\left|x-1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x=-6\end{matrix}\right.\)
g, \(\left|-x+4\right|=\left|-3x-8\right|\)
\(\Rightarrow\left\{{}\begin{matrix}-x+4=3x+8\\-x+4=-3x-8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-4x=4\\2x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
h, \(\left|x+4\right|=\left|-3-8\right|\)
\(\Rightarrow\left|x+4\right|=\left|-11\right|=11\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=-11\\x+4=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-15\\x=7\end{matrix}\right.\)
Chúc bạn học tốt!!!
NT
1
28 tháng 7 2018
a, x + \(\dfrac{1}{5}\) = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{5}\)
x = \(\dfrac{19}{30}\)
b, x - \(\dfrac{3}{5}=\dfrac{6}{7}\)
x = \(\dfrac{6}{7}+\dfrac{3}{5}\)
x = \(1\dfrac{16}{35}\)
c, - x - \(\dfrac{7}{5}=\dfrac{-8}{9}\)
- x = \(\dfrac{-8}{9}+\dfrac{7}{5}\)
x = \(\dfrac{23}{45}\)
d, \(\dfrac{3}{8}-x=\dfrac{2}{3}\)
\(x=\dfrac{3}{8}-\dfrac{2}{3}\)
\(x=\dfrac{-7}{24}\)
e, \(|x-\dfrac{8}{9}|=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{8}{9}\)
\(x=\dfrac{-11}{9}\)
3 tháng 8 2018
Hiệu giữa SBT mới và cũ là:
353 – 23 = 330
Hiệu số phần bằng nhau là:
3-1 = 2 phần
Số bị trừ cũ là: 330 : 2 = 165
Số trừ cũ là : 165- 23 = 142
3 tháng 6 2021
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
DL
1
21 tháng 5 2022
a: -3x+4+5x=-10-x
=>2x+4=-x-10
=>3x=-14
hay x=-14/3
b: \(-x+1=-3x-8\)
=>-x+3x=-8-1
=>2x=-9
hay x=-9/2
c: \(8-\left(x-1\right)=10+\left(x+5\right)\)
=>x+15=8-x+1
=>x+15=9-x
=>2x=-6
hay x=-3
d: \(100+\left(x+7\right)-\left(-x+3\right)=8+\left(x+100\right)\)
=>x+7+x-3=8+x
=>2x+4-x-8=0
=>x=4
14 tháng 6 2023
a: =7/8:(2/9-18+1/36)-5/12
=-7/142-5/12=-397/852
b: =3/7(4/9+5/9:6/12)=2/3
c: =5^8(16/31-47/31)+1/3=-5^8+1/3
d: =7/2(3/8+5/8:4/15)=609/64
ĐKXĐ: x<>1
\(\dfrac{x-1}{8}=\dfrac{8}{x-1}\)
=>\(\left(x-1\right)\cdot\left(x-1\right)=8\cdot8\)
=>\(\left(x-1\right)^2=64\)
=>\(\left[{}\begin{matrix}x-1=8\\x-1=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8+1=9\left(nhận\right)\\x=-8+1=-7\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{x-1}{8}\) = \(\dfrac{8}{x-1}\) (đk \(x-1\ne0\) ⇒ \(x\ne\) 1)
(\(x-1\)).(\(x-1\)) = 8.8
(\(x-1\))2 = 82
\(\left[{}\begin{matrix}x-1=-8\\x-1=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+1\\x=8+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-7\\x=9\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-7; 9)