NH
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HH
18 tháng 6 2018
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
20 tháng 6 2017
2) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$
$=>\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$
$=>\dfrac{x+4}{2000}+\dfrac{2000}{2000}+\dfrac{x+3}{2001}+\dfrac{2001}{2001}=\dfrac{x+2}{2002}+\dfrac{2002}{2002}+\dfrac{x+1}{2003}+\dfrac{2003}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$
$=>(x+2004)(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}=0$
$=>x+2004=0$
$=>x=-2004$
20 tháng 6 2017
3) Ta có : $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$
$=>A=\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$
$=>A>\dfrac{7}{12}(1)$
Ta lại có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}<(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
$=>A<\dfrac{5}{6}(2)$
Từ (1)(2) => đpcm.
12 tháng 1 2018
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
22 tháng 10 2017
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!
14 tháng 8 2022
a: =>||12x-1/2|-2|=-2/3x3/4=-6/12=-1/2(loại)
b: =>2/3-1/3x-1/2+2/3x=2x+2/3
=>-5/3x=1/2
=>x=-1/2:5/3=-1/2x3/5=-3/10
c: =>|3/2x+1/4|=2+3/4=11/4
=>3/2x+1/4=11/4 hoặc 3/2x+1/4=-11/4
=>3/2x=5/2 hoặc 3/2x=-3
=>x=3/5 hoặc x=-3:3/2=-2
CN
3 tháng 1 2018
\(\left(\dfrac{2}{5}+\dfrac{1}{4}\right)\left(\dfrac{-x}{3}+\dfrac{1}{2}\right)+\dfrac{2}{3}=\dfrac{7}{4}\)
⇔ \(\dfrac{13}{20}\left(\dfrac{-x}{3}+\dfrac{1}{2}\right)=\dfrac{7}{4}-\dfrac{2}{3}\)
⇔ \(\dfrac{13}{20}\left(\dfrac{-x}{3}+\dfrac{1}{2}\right)=\dfrac{13}{12}\)
⇔ \(\dfrac{-x}{3}+\dfrac{1}{2}=\dfrac{13}{12}:\dfrac{13}{20}\)
⇔ \(\dfrac{-x}{3}+\dfrac{1}{2}=\dfrac{5}{3}\)
⇔ \(\dfrac{-x}{3}=\dfrac{5}{3}-\dfrac{1}{2}\)
⇔ \(\dfrac{-x}{3}=\dfrac{7}{6}\)
⇔ \(\dfrac{-2x}{6}=\dfrac{7}{6}\)
⇔ -2x = 7
⇔ \(x=-\dfrac{7}{2}\)
b) \(\left(\dfrac{3}{5}-\dfrac{x}{2}\right)\left(\dfrac{4}{5}+\dfrac{1}{2}\right)-\dfrac{4}{3}=\dfrac{3}{2}\)
⇔ \(\dfrac{13}{10}\left(\dfrac{3}{5}-\dfrac{x}{2}\right)=\dfrac{3}{2}+\dfrac{4}{3}\)
⇔ \(\dfrac{13}{10}\left(\dfrac{3}{5}-\dfrac{x}{2}\right)=\dfrac{17}{6}\)
⇔ \(\dfrac{3}{5}-\dfrac{x}{2}=\dfrac{17}{6}:\dfrac{13}{10}\)
⇔ \(\dfrac{3}{5}-\dfrac{x}{2}=\dfrac{85}{39}\)
⇔ \(\dfrac{x}{2}=\dfrac{3}{5}-\dfrac{85}{39}\)
⇔ \(\dfrac{x}{2}=\dfrac{-308}{195}\)
⇔ \(\dfrac{195x}{390}=\dfrac{-616}{390}\)
⇔ 195x = -616
⇔ \(x=\dfrac{-616}{195}\)
2 tháng 8 2018
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
5 tháng 8 2018
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
13 tháng 7 2017
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
3/4 - ( x + 1/2 ) = 1/4
x + 1/2 = 3/4 - 1/4
x + 1/2 = 1/2 ( rút gọn của 2/4 )
x = 1/2 - 1/2
x = 0
x=0