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a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)
\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)
Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)
b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)
Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)
g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)
\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)
h, Vì E < 1 nên:
\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)
Vậy E = F
a)73+86+968+914+3032
= (86 + 914) +(968 + 3032) + 73
= 1000 + 4000 + 73
= 5073
b)341.67+341.16+659.83
=341 (67 + 16) + 659 * 83
= 341 * 83 + 659 * 83
= (341+ 659) * 83
= 1000 * 83
= 830000
c)252-84:21+7
= 252 - 2 +7
=257
d)4.8.25.125.27
= (8 * 125) (4* 25) * 27
= 1000 * 100 * 27
= 2700000
e)34+35+36+37-24-25-26-27
= (34-24) + (35-25) + (36-26) + (37-27)
= 10 + 10 + 10 + 10
= 40
g)1-2+3-4+...-98+99
= (1 + 99) + (-2-98) + .....+ (-48 - 52) + (49 + 51) - 50
= 100 - 50
= 50
h)1-4+7-10+...-100+103
Bài 2:Tìm số nguyên x,biết:
l)3636:(12.x-91)=36
3636 : (12x - 91) = 36
12x - 91 = 101
12x = 192
x= 16
m)(x:23+45).67=8911
x: 23 + 45 = 133
x : 23 = 88
x= 88/23
n)[(6.x-39):7].4=12
6x - 39 = 3
6x = 42
x=7
p)128-3(x+4)=23
3(x+4) = 105
x+4= 35
x= 31
q)[(4.x+28).3+55]:5=35
(4x + 28)*3 +55 = 175
(4x + 28) * 3 = 120
4x + 28 = 40
4x = 12
x=3
r)123-5.(x+4)=38
5(x+4) = 85
x+4 = 17
x= 13
t)11-(53+x)=97
53 + x = -86
x= -139
i)23.75+25.23+180
= 23(75+25) + 180
= 2300 + 180
= 2480
a)A có: (26-12)/2+1=8(số hạng)
A=(26+12)*8/2=152
b)B=1+(-2)+3+(-4)+...+2003+(-2004)+2005=(-2+1)+(-4+3)+...+(-2004+2003)+2005=(-1)+(-1)+...+(-1)+2005=(-1)*1002+2005=-1002+2005=1003
haizzz, còn lại lười làm quá