Tính A= \(\frac{-17,5+\frac{5}{3}+2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)
Tính hợp lý:
a) \(\left(4\frac{2}{3}-2\frac{2}{5}+7\frac{7}{13}\right)-\left(3\frac{5}{23}-6\frac{6}{13}\right)\)
b) \(\frac{-17,5+\frac{5}{3}+2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)
a) \(\left(4\frac{2}{3}-2\frac{2}{5}+7\frac{7}{13}\right)-\left(3\frac{5}{23}-6\frac{6}{13}\right)\)
= \(4\frac{2}{3}-2\frac{2}{5}+7\frac{7}{13}-3\frac{5}{23}+6\frac{6}{13}\)
= \(\left(4\frac{2}{3}-2\frac{2}{5}\right)+\left(7\frac{7}{13}+6\frac{6}{13}\right)-3\frac{5}{23}\)
= \(2\frac{4}{15}+14-3\frac{5}{23}\)
= \(16\frac{4}{15}-3\frac{5}{23}\)
= \(13\frac{17}{345}\)
XIN LỖI NHA, MÌNH KO LÀM ĐƯỢC XONG CÂU b) , MONG BẠN THÔNG CẢM CHO MÌNH, KHI NÀO MÌNH NGHĨ RA CÂU b) MÌNH SẼ TRẢ LỜI
Giúp mình nhé =))
\(C=\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)
Ta có :\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}=\frac{-17,5+\frac{5}{3}-\frac{15}{7}}{7-\frac{2}{3}+\frac{6}{7}}=\frac{-2,5\left(.7-\frac{2}{3}+\frac{6}{7}\right)}{7-\frac{2}{3}+\frac{6}{7}}=-2,5\)
C=\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)
=\(\frac{\frac{-367,5}{21}+\frac{35}{21}-\frac{45}{21}}{\frac{147}{21}-\frac{14}{21}+\frac{18}{21}}\)
=\(\frac{\frac{-377,5}{21}}{\frac{151}{21}}\)
=\(-\frac{5}{2}\)
\(C=\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7,0-\frac{2}{3}+\frac{6}{7}}\)
Tính hợp lí nha
Đúng + nhanh => mik tik
tính hợp lí
1, \(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}\)
2, \(5,25+3\frac{4}{9}+2\frac{1}{27}+4,75+3\frac{1}{4}+2+\frac{3}{4}\)
3, \(12\frac{3}{5}+\frac{17}{24}+\frac{7}{5}\)
4, \(4\frac{2}{3}+2\frac{1}{5}+2\frac{2}{3}+3\frac{3}{5}\)
Tính nhanh:
\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}\)
\(\frac{-17,5+\frac{5}{3}-2\frac{1}{7}}{7-\frac{2}{3}+\frac{6}{7}}=\frac{\frac{-35}{2}+\frac{5}{3}-\frac{15}{7}}{\frac{14}{2}-\frac{2}{3}+\frac{6}{7}}=\frac{-5.\left(\frac{7}{2}-\frac{1}{3}+\frac{3}{7}\right)}{2.\left(\frac{7}{2}-\frac{1}{3}+\frac{3}{7}\right)}=\frac{-5}{2}=-2,5\)
tính :
a) \(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}:\frac{-1}{7}\)
b) \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c) \(\frac{\left(13\frac{1}{4}-2\frac{5}{7}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
1. Tính :
a.\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
b.\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{3}{7}\right)\times......\times\left(1-\frac{10}{7}\right)\)
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
= \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
= \(\frac{2}{3}+\frac{1}{11}\)
= \(\frac{25}{33}\)
tính :
a) \(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}:\frac{-1}{7}\)
b) \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c) \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)
\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)
\(\frac{100}{-\frac{100}{41}}=-41\)
a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)
\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)
\(7:-\frac{1}{7}=-49\)
b. \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(-\frac{25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)
\(\frac{35}{6}:-\frac{43}{42}+\frac{15}{2}\)
\(-\frac{245}{43}+\frac{15}{2}=\frac{155}{86}\)
tính
2,\(A=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
3,\(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{1\frac{1}{6}-0,875+0,7}+\frac{6}{7}\)