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\(A=\frac{79}{1999}+\frac{191}{1998}+\frac{947}{1997}+\frac{673}{1998}+\frac{110}{1999}\)
\(A=\left(\frac{79}{1999}+\frac{110}{1999}\right)+\left(\frac{191}{1998}+\frac{673}{1998}\right)+\frac{947}{1997}\)
\(A=\frac{189}{1999}+\frac{16}{37}+\frac{947}{1997}\)
=4,034224056 mình cũng ko chắc nữa nhưng tịk giúp mình nha
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)
\(=\frac{1}{2000}\)
\(C=\frac{1999^{2000}+1}{1999^{1999}+1}< \frac{1999^{1999}+1+1998}{1999^{2000}+1+1998}\)
\(=\frac{1999^{1999}+1999}{1999^{2000}+1999}\)
\(=\frac{1999\cdot(1999^{1998}+1)}{1999\cdot(1999^{1999}+1)}\)
\(=\frac{1999^{1999}+1}{1999^{1998}+1}=D\)
Vậy...
C= 1999 1999 +1 1999 2000 +1 < 1999 2000 +1+1998 1999 1999 +1+1998 = 199 9 1999 + 1999 199 9 2000 + 1999 = 1999 2000 +1999 1999 1999 +1999 = 1999 ⋅ ( 199 9 1998 + 1 ) 1999 ⋅ ( 199 9 1999 + 1 ) = 1999⋅(1999 1999 +1) 1999⋅(1999 1998 +1) = 199 9 1999 + 1 199 9 1998 + 1 = D = 1999 1998 +1 1999 1999 +1 =D Vậy...
đáp án là < 1