\(\dfrac{3x}{20}\)=\(\dfrac{21}{5}\)
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16 tháng 4 2022
5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)
b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)
c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)
23 tháng 3 2021
Ta có: \(A=\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
\(=1-1+\dfrac{-5}{7}\)
\(=\dfrac{-5}{7}\)
30 tháng 5 2017
ai trả lời đúng được k nha các bạn. Giúp mình đi, mình đang vội
30 tháng 5 2017
Bn tham khảo ở đây nha trang nhí nhảnh
Câu hỏi của Tran Mai - Toán lớp 6 | Học trực tuyến
O
5 tháng 3 2022
\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)
\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)
\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)
\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)
5 tháng 3 2022
\(\dfrac{-1}{6}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-11}{15}\)
\(\dfrac{-9}{8}\)
28 tháng 7 2023
1: =5/13+8/13-20/41-21/41-5/17
=1-1-5/17=-5/17
2: =1/5+4/5-2/9-7/9+16/17
=16/17+1-1=16/17
3: =1/3-1/5+1/5-1/7+...+1/99-1/101
=1/3-1/101
=98/303
26 tháng 11 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)(nhận)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+2=1\\3+\dfrac{2}{x+y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x+y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=-1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)(nhận)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}y< >2x\\y< >-x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=6\\2x-y=3\end{matrix}\right.\)
=>x=2 và y=2x-3=4-3=1(nhận)
d:ĐKXĐ: \(\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)\)
e:
ĐKXĐ: \(x\ne\pm2y\)
\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{6}{x+2y}=5\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)\)
14 tháng 1 2023
c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)
=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)
=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12
=>-16x-1=-15x+12
=>-x=13
=>x=-13
3x/20 = 21/5
3x.5 = 21.20
15x = 420
x = 420 : 15
x = 28