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B=2+1/1.2+2+1/2.3+........+2+1/9.10
B=2.9+1/1.2+1/2.3+........+1/9.10
B=18+9/10
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=2\)
Đặt A=............(đề bài)
=>2A=\(2\left(\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\right)\)
=>2A=\(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)
=>2A=\(\left(\frac{2}{1.3}+...+\frac{2}{7.9}\right)+\left(\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
=>2A=\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
=>2A=\(\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\)
=>2A=\(\frac{8}{9}+\frac{2}{5}\)
=>2A=\(\frac{58}{45}\)
=>A=\(\frac{58}{45}:2\)
=>A=\(\frac{29}{45}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\dfrac{1}{7}A=\dfrac{1}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)
\(=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(=\dfrac{7-2}{2.7}+\dfrac{11-7}{7.11}+\dfrac{14-11}{11.14}+\dfrac{15-14}{14.15}+\dfrac{28-15}{15.28}\)
\(=\dfrac{7}{2.7}-\dfrac{2}{2.7}+\dfrac{11}{7.11}-\dfrac{7}{7.11}+\dfrac{14}{11.14}-\dfrac{11}{11.14}+\dfrac{15}{14.15}-\dfrac{14}{14.15}+\dfrac{28}{15.28}-\dfrac{15}{15.28}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)
\(=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{14}{28}-\dfrac{1}{28}=\dfrac{13}{28}\)
\(A=\dfrac{13}{28}\div\dfrac{1}{7}=\dfrac{13}{4}\)
Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\Rightarrow\dfrac{1}{7}.A=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(\Rightarrow\dfrac{1}{7}.A=\left(\dfrac{1}{2}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{14}\right)+\left(\dfrac{1}{14}-\dfrac{1}{15}\right)+\left(\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(\Rightarrow\dfrac{1}{7}.A=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)
\(\Leftrightarrow A=\dfrac{13}{4}\)
Vậy...................
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{\left(3.2^2.2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3^2.2}{11-2}\)
\(=2\)
= \(\dfrac{12,48:\dfrac{1}{2}.6,25.4.2}{2.3,12.1,25:\dfrac{1}{4}.10}\) ( dấu (.) là dấu x nha )
= \(\dfrac{12,48.2.6,25.4.2}{2.3,12.1,25.4.10}\)
(+) ta thấy 12,48 chia hết cho 3,12 =4
(+)2 v/s 2 chia hết loại
(+) 6,25 :1,25=5
(+)4v/s4 chia hết loại
(+)10: 2= 5
sau khi rút gọn ta có p/s sau:
= \(\dfrac{4.1.5.1.1}{1.1.1.1.5}\)= 4 ( 5: 5 = 1 loại )
vậy kết quả sẽ bằng 4 nha bn!!
thanks