1)Tính:
E=1/(2*9)+1/(9*7)+1/(7*19)+...+1/(252*509)
Tính nhanh : 1/(2*9) + 1/(9*7) + 1/(7*19) +..............+ 1/(252*509)
Tính: A= 1/(2*9) + 1/(9*7) + 1/(7*19)+...+ 1/(252*509)
Tính
E=1/2×9 +1/9×7 +/7×19 +...+1/252×509
Tính
A=1/2×9+1/9×7+1/7×19+...+1/252×509
B=1/10×9+1/18×13+1/26×17+...+1/802×405
C=2/4×7-3/5×9+2/7×10-3/9×13+...+2/301×304-3/401×405
B = \(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
B = \(\frac{2}{10.18}+\frac{2}{18.26}+\frac{2}{26.34}+...+\frac{2}{802.810}\)
B = \(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{8}{81}\)
B = \(\frac{2}{81}\)
Bài 1;tính
A=1/2*9+1/9*7+1/7*19+....+1/252*509
B=1/10*9+1/18*13+/26*17+....+1/802*405
C=2/4*7-3/5*9+2/7*10-3/9*13+...+2/301*304-3/401*405
Bài 2
Cho S=1/5^2+1/9^2+...+1/409^2
Chứng minh S<1/12
4S = 4/(5x5) + 4/(9x9) + … + 1/(409x409)
Ta thấy:
4/(5x5) < 4/(3x7) = 1/3 – 1/7
4/(9x9) < 4/(7x11) = 1/7 – 1/11
…………
4/(409x409) < 4/(407x411) = 1/407 – 1/411
Mà :
4/(3x7) + 4/(7x11) + …. + 4/(407x411) = 1/3 – 1/411 = 136/411
4S < 136/411
S < 34/411 < 34/408 = 1/12
Hay S < 1/12
1) -4\7 - 11\19 + 13\19 . -3\7 + 2\19 : -7\4
2) ( -4\9 + 3\5 ) : 1\1\5 + ( 1\5 - 5\9) : 1\1\5
3) 4\5 - ( -2\7) -7\10
4)2\7 - ( -13\15 + 4\9) - ( 5\9 - 2\15 )
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
Bài 1 Tìm X
X+X×1/2:2/9+X:2/7=252
x+x*\(\frac{1}{2}\):\(\frac{2}{9}\)+x:\(\frac{2}{7}\)=252
x+x*\(\frac{9}{4}\)+x*\(\frac{7}{2}\)=252
x*(1+\(\frac{9}{4}\)+\(\frac{7}{2}\))=252
x*\(\frac{27}{4}\)=252
x=252:\(\frac{27}{4}\)
x=\(\frac{112}{3}\)
Vậy x=\(\frac{112}{3}\)
Học tốt
\(x+x\times\frac{1}{2}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(x\times1+x\times\frac{9}{4}+x\times\frac{7}{2}=252\)
\(x\times\left(1+\frac{9}{4}+\frac{7}{2}\right)=252\)
\(x\times\frac{17}{4}=252\)
\(x=252\div\frac{17}{4}\)
\(x=252\times\frac{4}{17}\)
\(x=\frac{1008}{17}\)
Chắc sai =]]
y+y*1/3:2/9+y:2/7=252
Tính
a) 4/3*7 + 4/7*11 + 4/11*15 + 4/15*19 + 4/19*23 + 4/23*27
b) 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 +1/6*7
c) 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13 + 2/1*2 + 2/2*3 + 2/3*4 + 2/8*9 + 2/9*10
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)