Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
Giải:
a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{11}\)
\(2x=\dfrac{16}{11}-\dfrac{7}{2}\)
\(2x=\dfrac{-45}{22}\)
\(x=\dfrac{-45}{22}:2\)
\(x=\dfrac{-45}{44}\)
b) \(3-\left(17-x\right)=-12\)
\(3-17+x=-12\)
\(x=-12-3+17\)
\(x=2\)
c) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}:\dfrac{2}{3}\)
\(x=\dfrac{-3}{5}\)
d) \(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\)
\(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)
Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\)
e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\)
\(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\)
\(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\)
\(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\)
\(-0,6x=\dfrac{-7}{18}\)
\(x=\dfrac{-7}{18}:-0.6\)
\(x=\dfrac{35}{54}\)
f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\)
\(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\)
\(\dfrac{3}{5}.x=\dfrac{13}{9}\)
\(x=\dfrac{13}{9}:\dfrac{3}{5}\)
\(x=\dfrac{65}{27}\)
Chúc bạn học tốt!
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: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5}
=0+\dfrac{3}{5}=\dfrac{3}{5}\)
b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)
c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)
Bài 1:
a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)
b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)
\(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
a:=>0,75x-x+1,25x=0,2
=>x=0,2
b: =>1/3-x=-3/6+4/6=1/6
=>x=1/3-1/6=1/6
c: =>(x-1)/45=-6/30=-1/5
=>x-1=-9
=>x=-8
d: =>(2/5x-1)=-5/7
=>2/5x=2/7
=>x=5/7
a) \(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot1\dfrac{2}{3}=1\dfrac{1}{4}\)
\(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot\dfrac{5}{3}=\dfrac{5}{4}\)
\(\dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{5}{4}:\dfrac{5}{3}\\ \dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{3}{4}\\ \dfrac{1}{5}x=\dfrac{3}{20}-\dfrac{3}{4}\\ \dfrac{1}{5}x=-\dfrac{3}{5}\\ x=-\dfrac{3}{5}:\dfrac{1}{5}\\ x=-3\)
b) \(\dfrac{-2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ \dfrac{-2}{3}x=\dfrac{1}{10}\\ x=\dfrac{1}{10}:\dfrac{-2}{3}\\ x=-\dfrac{3}{20}\)
c) \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-7\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \dfrac{1}{3}\left(2x-7\right)=-\dfrac{13}{6}\\ 2x-7=-\dfrac{13}{6}:\dfrac{1}{3}\\ 2x-7=-\dfrac{13}{2}\\ 2x=-\dfrac{13}{2}+7\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{4}\)