Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
Giải:
a) \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{-1}{2}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{6}:\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{5}{18}\)
Vậy \(x=\dfrac{5}{18}\).
b) \(\left(\dfrac{1}{2}-x\right).\dfrac{2}{3}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{2}-x=\dfrac{1}{8}:\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{2}-x=\dfrac{3}{16}\)
\(\Leftrightarrow x=\dfrac{1}{2}-\dfrac{3}{16}\)
\(\Leftrightarrow x=\dfrac{5}{16}\)
Vậy \(x=\dfrac{5}{16}\).
c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{3}{4}+\dfrac{1}{2}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{47}{28}\\2x=-\dfrac{23}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{47}{56}\\x=-\dfrac{23}{56}\end{matrix}\right.\)
Vậy \(x=\dfrac{47}{56}\) hoặc \(x=-\dfrac{23}{56}\).
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(\Leftrightarrow2\left(2x+1\right)=3\left(x-5\right)\)
\(\Leftrightarrow4x+2=3x-15\)
\(\Leftrightarrow4x-3x=-15-2\)
\(\Leftrightarrow x=-17\)
Vậy \(x=-17\).
Chúc bạn học tốt!!!
a. \(\dfrac{3}{5}x-\dfrac{2}{3}=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{5}{18}\)
b) \(\left(\dfrac{1}{2}-x\right).\dfrac{2}{3}=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{5}{16}\)
c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{47}{56}\\x=\dfrac{-23}{56}\end{matrix}\right.\)
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow x=-17\).
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)