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Ta có :2013A=2013.2013^2012+1/2013^2013+1=2013^2013+2013/2013^2013+1=[2013^2013+1]+2012/2013^2013+1=1+2012/2013^2013+1
2013B=2013.2013^2013+1/2013^2014+1=2013^2014+2013/2014^2014+1=[2013+1]+2012/2013^2014+1=1+2012/2013^2014+1
Ta thấy:1+2012/2013^2013+1>1+2013/2013^2014+1 suy ra 2015A>2015B
a) \(15+2\left|x\right|=-3\\ \\ < =>2\left|x\right|=15-\left(-3\right)\\ < =>2\left|x\right|=18\\ =>\left|x\right|=\frac{18}{2}=9\\ =>x=9hoặcx=-9\)
b) \(\left|x-2\right|=7\\ < =>x-2=7hoặcx-2=-7\\ =>x=9hoặcx=-5\)
c) \(100-4.x^2=224\\ < =>4.x^2=100-224=-124\\ < =>x^2=-\frac{124}{4}=-31\\ Mà:x^2\ge0\\ =>xkhôngcógiátrịnàothoảmãn\)
d)\(2x-\frac{9}{240}=\frac{39}{80}\\ < =>2x-\frac{3}{80}=\frac{39}{80}\\ =>2x=\frac{39}{80}+\frac{3}{80}=\frac{21}{40}\\ =>x=\frac{\frac{21}{40}}{2}=\frac{21}{80}\)
A=33. \(\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)...\left(1-\frac{2}{99}\right)\)
A=33.\(\frac{1}{3}.\frac{3}{5}....\frac{97}{99}\)
A=33.\(\frac{1}{99}\)
A=\(\frac{33}{99}=\frac{1}{3}\)
a , \(A=\frac{19^{30}+1}{19^{31}+1}\Rightarrow19A=\frac{19^{31}+19}{19^{31}+1}=\frac{19^{31}+1+18}{19^{31}+1}=1+\frac{18}{19^{31}+1}\)
\(B=\frac{19^{31}+1}{19^{32}+1}\Rightarrow19B=\frac{19^{32}+19}{19^{32}+1}=\frac{19^{32}+1+18}{19^{32}+1}=1+\frac{18}{19^{32}+1}\)
Vì \(19A< 19B\Leftrightarrow A< B\)
b, câu b tương tự nha
a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
b: \(\left(\dfrac{1}{32}\right)^7=\left(\dfrac{1}{2}\right)^{35}\)
\(\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{36}\)
mà 35<36
nên \(\left(\dfrac{1}{32}\right)^7< \left(\dfrac{1}{16}\right)^9\)
\(a=\frac{2011\cdot2010+2000}{2010\cdot2012-10}=\frac{2011\cdot2010+2000}{2010\cdot2011+2010-10}=\frac{2011\cdot2010+2000}{2010\cdot2011+2000}=1\)
\(b=\frac{2013}{2012}>1\) (vì 2013 > 2012)
\(\Rightarrow\) a < b