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6 tháng 8 2019
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)=\frac{1}{2}.\frac{7-1}{21}=\frac{1}{7}\)
XO
6 tháng 8 2019
\(\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{19\times21}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{19\times21}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{2}\times\frac{2}{7}\)
\(=\frac{1}{7}\)
TN
0
TK
2
KT
20 tháng 1 2018
a) \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Leftrightarrow\)\(A=2-\frac{1}{2^7}=\frac{255}{128}\)
KT
20 tháng 1 2018
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{2}{7}=\frac{1}{7}\)
9 tháng 10 2018
\(a)5\frac{3}{5}+1\frac{3}{4}+4\frac{1}{4}+3\frac{2}{5}\)
\(=\)\((5\frac{3}{5}+3\frac{2}{5})+(1\frac{3}{4}+4\frac{1}{4})\)
\(=[8+(\frac{3}{5}+\frac{2}{5})]+[5+(\frac{3}{4}+\frac{1}{4})]\)
\(=(8+1)+(5+1)\)
\(=9+6=15\)
\(b)\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=(\frac{3}{4}+\frac{1}{4})+(\frac{18}{21}+\frac{3}{21})+(\frac{19}{32}+\frac{13}{32})\)
\(=1+1+1=3\)
_Học tốt_
22 tháng 8 2018
A = 1 * 3 + 3 * 5 + ... + 21 * 23
A = 1 * (3+3 * 5 + ... + 21) * 23
A = 1 * 23
A = 23
5 tháng 6 2023
\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)
= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)
=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)
= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)
\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)
= \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)
= 1 + 1 + 1 = 3
\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)
= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!
5 tháng 6 2023
`1,`
`a,`
`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`
`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`
`= 7/19 \times (8/23 + 15/23 + 1) + 1`
`= 7/19 \times 2 + 1`
`=14/19 + 1`
`= 33/19`
`b,`
`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`
`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`
`= 0,75 + 1 + 1 + 0,25`
`= (0,75 + 0,25) + 1 + 1`
`= 1+1+1=3`
`c,`
`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`
`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)
`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}+\frac{1}{21.23}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}+\frac{2}{21.23}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{23}\right)\)
\(A=\frac{1}{2}.\frac{22}{23}\)
\(A=\frac{11}{23}\)