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Ta có: \(N=\dfrac{5}{3.5}-\dfrac{5}{5.7}-\dfrac{5}{7.9}-\dfrac{5}{9.11}-\dfrac{5}{11.13}-\dfrac{5}{13.15}
\)
\(\Rightarrow N=5\left(\dfrac{1}{3.5}-\dfrac{1}{5.7}-\dfrac{1}{7.9}-\dfrac{1}{9.11}-\dfrac{1}{11.13}-\dfrac{1}{13.15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{13}-\dfrac{1}{13}+\dfrac{1}{15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{15}\right)=\dfrac{5}{2}\left(\dfrac{5+1}{15}\right)=\dfrac{5}{2}.\dfrac{6}{15}\)
\(\Rightarrow N=\dfrac{3}{3}=1\)
1) P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15
P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)
P=1/3-1/15= 4/15
2) a/ 0,2:1+3/5+80%
= 2/10:8/5+8/10
= 2/10.5/8+8/10
= 1/8 + 4/5 = 5/40 + 32/40 = 37/40
b/ 0,5:5/4-2+1/5
= 5/10:5/4-11/5
= 5/10.4/5-11/5
=2/5-11/5 = -9/5
Đặt 2/5 ra ngoài rồi tách từng cặp phân số ra sau đó bn tự làm nhé!
A=\(\frac{5}{5.7}+\frac{5}{7.9}+.........+\frac{5}{59.61}\)
=\(\frac{5}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+.........+\frac{2}{59.61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...........+\frac{1}{59}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\frac{56}{305}\)
=\(\frac{28}{61}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+\frac{13-11}{11.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{5}{39}\)
a,\(\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{11.13}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{11}-\frac{1}{13}\)
=\(\frac{1}{3}-\frac{1}{13}\)
=\(\frac{10}{39}\)
b,Đặt A=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.............+\frac{1}{27.28.29.30}\)
3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...........+\frac{3}{27.28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.............+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{6}-\frac{1}{24360}\)
3A=\(\frac{1353}{8120}\)
A=\(\frac{451}{8120}\)
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1