Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(-\frac{2}{x}=\frac{y}{3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét bảng
| x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -1 |
| y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 6 |
Vậy.................
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Leftrightarrow2x-9=\frac{240.39}{80}\)
\(\Leftrightarrow2x-9=117\)
\(\Leftrightarrow2x=126\)
\(\Leftrightarrow x=63\)
Bài 1: Tìm x biết:
1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\) 2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)
x +\(\frac{7}{12}\)= \(\frac{15}{18}\) \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)
x = \(\frac{15}{18}\)- \(\frac{7}{12}\) \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)
x = \(\frac{1}{4}\) x = \(\frac{3}{10}\)
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}< x< \frac{-1}{2}+2+\frac{5}{2}\)
\(\Leftrightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}< x< \frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)
\(\Leftrightarrow\frac{-18}{6}< x< \frac{8}{2}\)
\(\Leftrightarrow-3< x< 4\)
\(\Leftrightarrow x\in\left\{-2;-1;0;1;2;3\right\}\)