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a, \(-\dfrac{1}{4}-\dfrac{3}{4}:x=-\dfrac{11}{36}\)
\(\Rightarrow\dfrac{3}{4}:x=-\dfrac{1}{4}-\left(-\dfrac{11}{36}\right)\)
\(\Rightarrow\dfrac{3}{4}:x=\dfrac{1}{18}\)
\(\Rightarrow x=\dfrac{3}{4}:\dfrac{1}{18}\)
\(\Leftrightarrow x=\dfrac{3}{4}.18=\dfrac{54}{4}\)
b, \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=-\dfrac{7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\left(-\dfrac{6}{5}+x\right):\left(-\dfrac{18}{5}\right)=-\dfrac{7}{4}+2\)
\(\left(-\dfrac{6}{5}+x\right):\left(-\dfrac{18}{5}\right)=\dfrac{1}{4}\)
\(\left(-\dfrac{6}{5}+x\right)=\dfrac{1}{4}.\dfrac{18}{5}\)
\(-\dfrac{6}{5}+x=\dfrac{9}{10}\)
\(\Rightarrow\)\(x=\dfrac{9}{10}-\left(-\dfrac{6}{5}\right)\)
\(\Leftrightarrow x=\dfrac{21}{10}\)
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15
=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)
=1 - 3/4
=1/4
b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2
=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)
=-8 +3/2 +1/3 -3/10
=-97/15
a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)
\(\dfrac{2}{5}+x=\dfrac{3}{12}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=\dfrac{5}{20}-\dfrac{8}{20}\)
\(x=\dfrac{-3}{20}\)
b)\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)
\(x=0:2\) \(x=0+\dfrac{1}{7}\)
\(x=0\) \(x=\dfrac{1}{7}\)
\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)
x= \(\dfrac{1.\left(-5\right)}{1.7}\)
\(x=\dfrac{-5}{7}\)
a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)
Lập bảng xét dấu:
| x | -2 | \(\dfrac{1}{2}\) | |||
| x + 2 | - | 0 | + | + | |
| x - \(\dfrac{1}{2}\) | - | - | 0 | + |
TH : Xét x < -2
Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)
-2x = 2\(\dfrac{1}{4}\)
=> x = \(-1\dfrac{1}{8}\) ( loại )
TH 2: \(-2\le x< \dfrac{1}{2}\)
Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
=> \(2,5=\dfrac{3}{4}\) ( loại )
TH3 : \(x\ge\dfrac{1}{2}\)
x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
2x + 1,5 = \(\dfrac{3}{4}\)
x = -0,375( loại )
vậy ....
b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)
c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)
TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)
\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)
TH2 : \(x-1< 0\Rightarrow x< 1\)
=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)
Vậy x = 1
a: =>x+2/5=11/12-2/3=11/12-8/12=3/12=1/4
=>x=1/4-2/5=5/20-8/20=-3/20
b: \(\Leftrightarrow x\cdot\dfrac{11}{4}=\dfrac{11}{7}:\dfrac{1}{100}=\dfrac{1100}{7}\)
=>x=1100/7:11/4=400/7
c: =>x=0 hoặc x-1/7=0
=>x=0 hoặc x=1/7
d: =>2x=608/15
=>x=304/15
a, \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{4}{5}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}\\ \Rightarrow x=\dfrac{3}{10}\cdot\dfrac{3}{2}\\ \Rightarrow x=\dfrac{9}{20}\)
b, \(x+\dfrac{1}{4}=\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{4}{3}-\dfrac{1}{4}\\ \Rightarrow x=\dfrac{13}{12}\)
c, \(\dfrac{3}{5}x-\dfrac{1}{2}=-\dfrac{1}{7}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{1}{7}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{5}x=-\dfrac{9}{14}\\ \Rightarrow x=-\dfrac{9}{14}\cdot\dfrac{5}{3}\\ \Rightarrow x=\dfrac{15}{14}\)
d, \(\left|x-\dfrac{4}{5}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=-\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}+\dfrac{4}{5}\\x=-\dfrac{3}{4}+\dfrac{4}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)
e, \(lxl\) là j mk ko hiểu!
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
1,\(\dfrac{-1}{4}-\dfrac{3}{4}:x=-\dfrac{11}{36}\)
\(-\dfrac{3}{4}:x=\left(-\dfrac{1}{4}\right)-\left(-\dfrac{11}{36}\right)\)
\(-\dfrac{3}{4}:x=\dfrac{1}{18}\)
\(x=\left(-\dfrac{3}{4}\right):\left(\dfrac{1}{18}\right)\)
\(x=\dfrac{27}{2}\)
2, \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)
\(\dfrac{3}{4}x=\dfrac{3}{7}+\dfrac{1}{2}\)
\(\dfrac{3}{4}x=\dfrac{13}{14}\)
\(x=\dfrac{13}{14}:\dfrac{3}{4}\)
\(x=\dfrac{26}{21}\)
1 , x = 36/11
2 , x = 26/21