Bài 1:
a) \(\left[\dfrac{3}{20}-\dfrac{1}{5}x\right].1\dfrac{2}{3}=1\dfrac{1}{4}\)
b)\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
c)\(\dfrac{-2}{3} \)-\(\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)
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AH
Akai Haruma
Giáo viên
25 tháng 7 2021
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
AH
Akai Haruma
Giáo viên
25 tháng 7 2021
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
5 tháng 3 2021
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
AH
Akai Haruma
Giáo viên
5 tháng 3 2021
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
29 tháng 1 2022
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
25 tháng 7 2017
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
22 tháng 6 2021
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!
26 tháng 9 2021
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
26 tháng 9 2021
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
8 tháng 12 2021
\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)
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) \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-7\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \dfrac{1}{3}\left(2x-7\right)=-\dfrac{13}{6}\\ 2x-7=-\dfrac{13}{6}:\dfrac{1}{3}\\ 2x-7=-\dfrac{13}{2}\\ 2x=-\dfrac{13}{2}+7\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{4}\)