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Bài 1:
a)12,5 x (-5/7) + 1,5 x (-5/7)
=-5/7*(12,5+1,5)
=-5/7*14
=-10
b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)
=-1/4*(68/11+42/11)
=-1/4*10
=-5/2
c tương tự
d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)
Bài 2:
a)Ta có:
2800=(28)100=256100
8200=(82)100=64100
Vì 256100>64100 =>2800>8200
b)Ta có:
1245=(123)15=172815
Vì 62515<172815 =>62515<1245
a,12,5x(-5/7)+1,5x(-5/7)
=-125/14+-15/14
=-10
2,2mu800>8 mu 200
6254 lon hon 12
\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)
\(2A=2+2^2+2^3+...+2^{51}\)
\(2A-A=A=2^{51}-2^0\)
\(B=5+5^2+5^3+...+5^{99}+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)
\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)
\(3C+C=4C=3^{2011}+3\)
\(C=\frac{3^{2011}+3}{4}\)
\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)
\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)
\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)
\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)
\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)
A=20+21+22+23+...++23+...+250250
2�=2+22+23+...+2512A=2+22+23+...+251
2�−�=�=251−202A−A=A=251−20
�=5+52+53+...+599+5100B=5+52+53+...+599+5100
5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101
5�−�=4�=5101−55B−B=4B=5101−5
�=5101−54B=45101−5
�=3−32+33−34+...+C=3−32+33−34+...+32007−32008+32009−3201032007−32008+32009−32010
3�=32−33+34−35+...−32008+32009−32010+320113C=32−33+34−35+...−32008+32009−32010+32011
3�+�=4�=32011+33C+C=4C=32011+3
�=32011+34C=432011+3
�100=5+5×9+5×92+5×93+...+5×999S100=5+5×9+5×92+5×93+...+5×999
�100=5×(1+9+92+93+...+999)S100=5×(1+9+92+93+...+999)
9�100=5×(9+92+93+...+999+9100)9S100=5×(9+92+93+...+999+9100)
9�100−�100=8�100=5×(9100−1)9S100−S100=8S100=5×(9100−1)
�100=5×(9100−1)8S100=85×(9100−1)
tính :
a)\(\frac{459}{987}\cdot\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2009}{2010}\)
\(=\frac{459}{987}\cdot\left(\frac{3}{12}+\frac{1}{12}-\frac{4}{12}\right)-\frac{2009}{2010}\)
\(=\frac{459}{987}\cdot0-\frac{2009}{2010}\)
\(=\frac{-2009}{2010}\)
b) \(\frac{-9}{7}-\frac{5}{7}\cdot\left[\left(\frac{-2}{3}\right)^2-1\right]\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\cdot\left[\frac{4}{9}-1\right]\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\cdot\frac{-5}{9}\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\)
\(=-2\)
tìm x:
a) \(\left(x-\frac{7}{18}\right)-\frac{15}{27}=\frac{-10}{27}\)
\(\left(x-\frac{7}{18}\right)=\frac{-10}{27}+\frac{15}{27}\)
\(\left(x-\frac{7}{18}\right)=\frac{5}{27}\)
\(x=\frac{5}{27}+\frac{7}{18}\)
\(\Rightarrow x=\frac{31}{54}\)
b) \(\left(3\frac{1}{2}-2x\right)\cdot\frac{11}{3}=7\frac{1}{3}\)
\(\left(\frac{7}{2}-2x\right)\cdot\frac{11}{3}=\frac{22}{3}\)
\(\left(\frac{7}{2}-2x\right)=\frac{22}{3}\div\frac{11}{3}\)
\(\left(\frac{7}{2}-2x\right)=2\)
\(2x=\frac{7}{2}-2\)
\(x=\frac{3}{2}\div2=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
1+5+5^2+..+5^8+5^9/1+5+5^2+...+5^8=5^9 (1)
1+3+3^2+...+3^9/1+3+3^2+...+3^8=3^9 (2)
Vậy(1)>(2)