12:2(1999+100)=
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\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(7215-7215\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).0\)
\(=0\)
\(1999.1999.1998-1998.1998.1999\)
\(=1999.1998.\left(1999-1998\right)\)
\(=1999.1998.1\)
Tham khảo nhé~

A = ( 1999/2011 - 2011/1999 ) - ( -12/1999 - 12/2011 )
=1999/2011-2011/1999+12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
=1-1
=0
Vậy A=0
B = 2/5 + [ 3/11 + (-2)/5 ]
=[2/5+(-2/5)]+3/11
=0+3/11
=3/11

a. Có: \(\frac{100^{101}+1}{100^{100}+1}>1\Rightarrow\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+\left(1+99\right)}{100^{100}+\left(1+99\right)}\)
\(\Rightarrow B>\frac{100^{101}+100}{100^{100}+100}\\ \Rightarrow B>\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\\ \Rightarrow B>\frac{100^{100}+1}{100^{99}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
b. Có: \(\frac{13^{16}+1}{13^{17}+1}< 0\Rightarrow\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+\left(1+12\right)}{13^{17}+\left(1+12\right)}\)
\(\Rightarrow B< \frac{13^{16}+13}{13^{17}+13}\\ \Rightarrow B< \frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\\ \Rightarrow B< \frac{13^{15}+1}{13^{16}+1}=A\\ \Leftrightarrow A>B\)
Vậy A > B
c. Có: \(\frac{1999^{2000}+1}{1999^{1999}+1}>1\Rightarrow\frac{1999^{2000}+1}{1999^{1999}+1}>\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\)
\(\Rightarrow B>\frac{1999^{2000}+1999}{1999^{1999}+1999}\\ \Rightarrow B>\frac{1999\left(1999^{1999}+1\right)}{1999\left(1999^{1998}+1\right)}\\ \Rightarrow B>\frac{1999^{1999}+1}{1999^{1998}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B

x = 5 y = 36
( 18 / 21 + 3 / 21 ) + ( 19 / 32 + 13 / 32 ) + ( 75 / 100 + 1 / 4 )
= 1 + 1 + 1 = 3
2008/2009 > 2010/2011 ; 1999/2001 < 12/11 ;2007/2006> 2006/2007 ; 101/100 >100/101
1.
x = 5
y = 36
2.
( 18 / 21 + 3 / 21 ) + ( 19 / 32 + 13 / 32 ) + ( 75 / 100 + 1 / 4 ) = 1 + 1 + 1 = 3
3.
> ; < ; > ; >

(1999/2011-2011/1999) - (-12/1999 - 12/2011)
= 1999/2011 - 2011/1999 + 12/1999 + 12/2011
= (1999/2011 + 12/2011) - 1999/1999 - 12/1999 + 12/1999
= ( 1-1 ) + ( -12/1999 + 12/1999)
= 0 + 0
= 0
(1999/2011-2011/1999)-(12/1999-12/2011)
=1999/2011-2011/1999-12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
=1-1
=0

\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\frac{2011}{2011}-\frac{1999}{1999}\)
\(=1-1=0\)
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\frac{2011}{2011}-\frac{1999}{1999}\)
\(=1-1=0\)
12 : 2(1999 + 100)
=12 : (3998 + 200)
=12: 4198
=0,0028585040495474
2105
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