Tính nhanh:
\(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
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Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
theo mk thỳ làm thế này :
\(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
= \(\left(1+2+3+4\right)+\left(\frac{4}{5}+\frac{4}{5}\right)+\left(\frac{5}{7}+\frac{5}{7}\right)\)
= \(10+\frac{8}{5}+\frac{10}{7}\)
r bn cộng vô là đc, mk nghĩ thế thôi, sai thỳ thông cảm.
=\(\frac{9}{5}+\frac{19}{7}+\frac{19}{5}+\frac{33}{7}\)
=\(\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
=\(\frac{28}{5}+\frac{52}{7}\)
=\(\frac{456}{35}\)
cho mk nha bn!!!!!!!!!!!!!!!!
Ta có:
G=\(3\frac{5}{7}+4\frac{3}{5}+1\frac{2}{7}\)\(+3\frac{1}{5}-\frac{4}{5}\)
=\(3+\frac{5}{7}+4+\frac{3}{5}+1+\frac{2}{7}+3+\frac{1}{5}-\frac{4}{5}\)
=\(\left(3+4+1+3\right)+\left(\frac{5}{7}+\frac{3}{5}+\frac{2}{7}+\frac{1}{5}-\frac{4}{5}\right)\)
=\(11+\left[\left(\frac{5}{7}+\frac{2}{7}\right)+\left(\frac{3}{5}+\frac{1}{5}-\frac{4}{5}\right)\right]\)
=\(11+\left[\frac{7}{7}+\frac{0}{5}\right]\)
=\(11+\left[1+0\right]\)
=\(11+1\)
=12
ĐÃ GIẢI CHI TIẾT RÙI NHÉ. **** NHA
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
\(\frac{5-\frac{5}{7}-\frac{5}{49}}{4-\frac{4}{7}-\frac{4}{49}}+\frac{1,5+75\%-\frac{3}{8}}{0.625-\frac{5}{2}-125\%}\)
\(=\frac{5.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{49}\right)}{4\cdot\left(\frac{1}{4}-\frac{1}{7}-\frac{1}{49}\right)}+\frac{\frac{3}{2}+\frac{3}{4}-\frac{3}{8}}{\frac{5}{8}-\frac{5}{2}-\frac{5}{4}}\)
\(=\frac{5}{4}+\frac{3\cdot\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}\right)}{5\cdot\left(\frac{1}{8}-\frac{1}{2}-\frac{1}{4}\right)}\)
\(=\frac{5}{4}+\left(-\frac{3}{5}\right)\)
\(=\frac{13}{20}\)
ta có: A = \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
=\(1+\frac{4}{5}+2+\frac{5}{7}+3+\frac{4}{5}+4+\frac{5}{7}\)
=\(\left(1+2+3+4\right)+\left(\frac{4}{5}+\frac{4}{5}+\frac{5}{7}+\frac{5}{7}\right)\)
= 10 + \(\frac{106}{35}\)
= 10 + \(3\frac{1}{35}\)
= \(13\frac{1}{35}\)