Tính nhanh: \(\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\frac{6}{4!+5!+6!}+\frac{7}{5!+6!+7!}+\frac{8}{6!+7!+8!}\)
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\(=\frac{5\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}{-4\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}:\frac{2\left(\frac{1}{3}-\frac{1}{12}+\frac{3}{7}\right)}{ }\)
MÃu thứ hai sao ý
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
a ) \(5\frac{3}{4}:3+2\frac{1}{4}.\frac{1}{3}-\frac{3}{8}=\frac{23}{4}:\frac{3}{1}+\frac{9}{4}.\frac{1}{3}=\frac{23}{12}+\frac{3}{4}=\frac{8}{3}\)
b ) \(\frac{3}{5}:\frac{5}{6}:\frac{6}{7}:\frac{7}{8}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{3.6.7.8}{5.5.6.7}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{24}{25}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{4049}{1400}\)
1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10
=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)
=9/10x(1/3+3/5+5/7+7/9)
9/10x(1/3+3/5)+(5/7+7/9)
=9/10x1/5+5/9
9/50+5/9
=10
Bn Long làm đúng rồi bn nguyễn kim arica cứ làm theo cách đó là được .
Bn nào thấy đúng thì ủng hộ nha .
Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)