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a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Ta có:
\(\frac{A}{5}=\frac{4}{35\cdot31}+\frac{6}{35\cdot41}+\frac{9}{50\cdot41}+\frac{7}{50\cdot57}=\frac{35-31}{35\cdot31}+\frac{41-35}{35\cdot41}+\frac{50-41}{50\cdot41}+\frac{57-50}{50\cdot57}\) ( Ps cuối là\(\frac{57-50}{50.57}\) nha).
\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\Rightarrow\frac{A}{5}=5\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Tương tự:
\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{38.31}+\frac{41-38}{38.41}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)
\(=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\)
\(\Rightarrow\frac{B}{2}=2\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Từ đó, suy ra:
\(\frac{A}{B}=\frac{5}{2}\)
Vậy \(\frac{A}{B}=\frac{5}{2}\)
\(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}\)
\(=\) \(\frac{1}{3}\)
\(\frac{1}{3}\)nha
~~ tk mk đi ~~
Ai tk mk mk tk lại ~~
kb nha ~ n_n
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
Mk chỉ biết câu a thôi
a) \(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}\)
= \(\frac{5\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}{15\cdot\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{11}\right)}\)
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
Chúc bạn học tốt
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!