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a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50
=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3
3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)
3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60
3A = -2.3.4 + 49.60.61
\(A=\frac{-2.3.4+49.60.61}{3}=59772\)
b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53
=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6
6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)
6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53
6B = 1.3 + 51.53.55
\(B=\frac{1.3+51.53.55}{6}=24778\)
cau c mk ko bk
d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
D = 30+31+32+33+34+35+36+37+38
=> 3D = 31+32+33+...+38+39
=> 3D - D = 39-30
2D = 39-1
\(D=\frac{3^9-1}{2}=9841\)
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
Bài 1.
A=100+98+96+...+2-97-95-...-1
A= 100 + (98-97) + (96-95) + ... +(2-1)
Từ 1 đến 98 có 98 số => có 98 : 2 cặp mà hiệu = 1
A = 100 + 49 x 1 = 149
Bài 2
B = 1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302
B = 1 + 2 + (302 - 300) + (301 - 299) + ... + (10 - 8) + (9-7) + (6-4) + (5-3)
Từ 3 đến 302 có 300 số => có 300 : 2 cặp hiệu = 2
B = 1 + 2 + 150 x 2 = 303
Ta thấy:
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{17\cdot19}\)
\(=\frac{1}{2}\cdot\frac{2}{3\cdot5}+\frac{1}{2}\cdot\frac{2}{5\cdot7}+...+\frac{1}{2}\cdot\frac{2}{17\cdot19}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\cdot\left[\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-...-\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{19}\right]\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-0-0-...-0-\frac{1}{19}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{19}{57}-\frac{3}{57}\right)\)
\(=\frac{1}{2}\cdot\frac{16}{57}\)
\(=\frac{1\cdot16}{2\cdot57}\)
\(=\frac{8}{57}\)
Khi đó:
\(\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{17\cdot19}\right)\cdot114-0,2\cdot\left(x-1\right)=10\)
\(\Rightarrow\frac{8}{57}\cdot114-\frac{1}{5}\cdot\left(x-1\right)=10\)
\(\Rightarrow8\cdot2-\left(x-1\right)\cdot\frac{1}{5}=10\)
\(\Rightarrow16-\frac{x-1}{5}=10\)
\(\Rightarrow\frac{x-1}{5}=16-10\)
\(\Rightarrow\frac{x-1}{5}=6\)
\(\Rightarrow x-1=6\cdot5\)
\(\Rightarrow x-1=30\)
\(\Rightarrow x=30+1\)
\(\Rightarrow x=31\)
Vậy x = 31