so sánh a và b biết \(A=\frac{11\cdot13\cdot15+33\cdot39\cdot45+55\cdot65\cdot75+99\cdot117\cdot135}{11\cdot13\cdot17+39\cdot45\cdot51+65\cdot75\cdot85+117\cdot135\cdot153}:B=\frac{1111}{1717}\)
1. Tính nhanh : A = 1/(1+2 ) + 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + .... + 1/(1+2+3+...+10)
2. So sánh A và B : A = ( 11 x 13 x 15 + 33 x 39 x 45 + 55 x 65 x 75 + 99 x 117 x 135 ) : ( 11 x 13x 17 + 39 x 45 x 57 + 65 x 75 x 85 + 117 x 135 x 153 ) B = 1111 : 1717
Bài 1:
\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)
\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=\frac{9}{11}\)
Bài 2 :
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)
\(=\frac{11}{17}\)
\(=\frac{1111}{1717}=B\)
Vậy \(A=B\)
Rút gọn phân thức
\(\frac{2^{35}\cdot45^{25}\cdot13^{22}\cdot35^{16}}{28^{17}\cdot9^{26}\cdot65^{22}\cdot25^9}\)
A=\(\frac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{4^{17}.7^{17}.9^{26}.13^{22}.5^{22}.5^9.5^9}=\frac{2^{35}.5^1}{4^{17}.7^1.9}=\frac{2^{35}.5}{2^{34}.7^1.9}\)= \(\frac{2.5}{7.9}=\frac{10}{63}\)
1. Tính nhanh :
A = 1/(1+2 ) + 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + .... + 1/(1+2+3+...+10)
2. So sánh A và B :
A = ( 11 x 13 x 15 + 33 x 39 x 45 + 55 x 65 x 75 + 99 x 117 x 135 ) : ( 11 x 13x 17 + 39 x 45 x 57 + 65 x 75 x 85 + 117 x 135 x 153 )
B = 1111 : 1717
1)\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
=> \(A
giải giùm mình bài này vớC= (67/111+2/33-15/117)x(1/3-1/4-1/12)
\(\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\cdot462-y=19\)
(2/143+2/195+2/255+2/323+2/399).462-y=19
=> 10/231.462-y=19
=> 20-y=19
=> y= 20-19=1
DUYỆT NHA "T I C K"
\(\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+....\frac{1}{41\cdot45}\)
Gọi A=\(\frac{1}{5.9}+\frac{1}{9.13}+....+\frac{1}{41.45}\)
=>4A=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{41.45}\)
4A=\(\frac{9-5}{5.9}+\frac{13-9}{9.13}+....+\frac{45-41}{41.45}\)
4A\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{41}-\frac{1}{45}\)
\(4A=\frac{1}{5}-\frac{1}{45}\)
\(4A=\frac{8}{45}\)
\(=>A=\frac{2}{45}\)
TTTTTTTTTa có:
\(S=\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{41.45}\Rightarrow4S=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......-\frac{1}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\frac{4}{45}\)
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot30}+...+\frac{1}{73\cdot75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...3+\frac{4}{2008\cdot2010}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
\(B=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
=
\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)
\(\Leftrightarrow\)\(x=\frac{-5}{11}\)
Vậy \(x=\frac{-5}{11}\)
Chúc bạn học tốt ~
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
=> 2.9 = x + 1
=> x + 1 = 18
=> x = 17
\(\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\cdot462-y=19\)
dấu chấm thành dấu nhân nha thông cảm
Tìm x thuộc N
\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(x+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}=\frac{2}{7}\)