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a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)
C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
Bài 2:b)Ta có:
D=(51*52*53*...*100):2^50.
=(51*53*55*...*99)*(52*54*56*...*100):2^50.
Khử 51*53*55*...*99 thì cần so sánh 1*3*5*...*41 với (52*54*56*...*100):2^50.
Lại có:
52*54*56*...*100:2^50=(52:2)*(54:2)*...*(100:2):(2^25) (vì 52;54;56;...;100 có 25 thừa số.
=26*27*28*...*50:2^25.
=(27*29*31*...*49)*(26*28*30*...*50):2^25
Khử với 1*3*5*...*49 thì cần so sánh 1*3*5*...*25 với (26*28*30*...*50):2^25.
Lại có:
26*28*30*...*50:2^25=(26:2)*(28:2)*(30:2)*...*(50:2):2^12(vì 26;28;30;...;50 có 13 thừa số).
=13*14*15*...*25:2^12.
=(13*15*17*19*21*23*25)*(14*16*18*20*22*24):2^12.
Khử với 1*3*5*...*25 thì cần so sánh 1*3*5*7*9*11 với (14*16*18*20*22*24):2^12.
Giờ số nhỏ rồi bấm máy tính so sánh là được.\
=>C=D.
Vậy C=D.
mấy câu kia dễ rồi tự l;àm nha mk nhắc câu khó thôi.
tk cho mk nha các bn.
-chúc ai tk mk học giỏi-
1/
a, x + (x+1) + (x+2) +...+ (x+100) = 2029099
(x+x+x+...+x) + (1+2+...+100) = 2029099
2011x + 2021055 = 2029099
2011x = 2029099 - 2021055
2011x = 8044
x = 8044 : 2011
x = 4
b, 2+4+6+....+2x = 210
=> 2(1+2+3+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x+1) = 14.15
=> x = 14
2/
a, Vì B < 1
\(\Rightarrow B< \frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}=\frac{2009^{2009}+2009}{2009^{2010}+2009}=\frac{2009\left(2009^{2008}+1\right)}{2009\left(2009^{2009}+1\right)}=\frac{2009^{2008}+1}{2009^{2009}+1}\)= A
Vậy A > B
b, Ta có:
\(D=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)
\(=\frac{\left(51.52.53....100\right)\left(1.2.3.4....50\right)}{2^{50}.\left(1.2.3.4....50\right)}\)
\(=\frac{1.2.3.4.5.6.....100}{\left(2.1\right)\left(2.2\right).\left(2.3\right).....\left(2.50\right)}\)
\(=\frac{1.2.3.4.5.6......100}{2.4.6........100}=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6....100}\)
\(=1.3.5....99=C\)
Vậy C = D
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)