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\(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+98\right)}{1.98+2.97+3.96+...+97.2+98.1}\)
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.98+2.97+3.96+...+98.1}=1\)
C= \(\frac{49}{200}\)
D= \(\frac{33}{100}\)
Chúc bạn Hk tốt!!!!
C =1/2*4+1/4*6+1/6*8+...+1/98*100
2xC=2/2*4+2/4*6+2/6*8+...+2/98*100
2xC=1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100
2xC=1/2-1/100
2xC=49/100
C=49/100:2
C=49/200
Ý B làm tương tự nhưng nhưng cả 2 vế với 3
nha. ^_^ ^_^ ^_^
Ta có:(1+2+3+4+5+...+97+98+99)x100/2x50:2500
=[(99+1)x99:2)]x50x50:2500
=4950x2500:2500
=4950x1
=4950
\(1+\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}\)
\(=1+1+\frac{1}{98}-\left(1+\frac{1}{97}\right)+\frac{1}{97}-\frac{1}{98}\)
\(=1+1+\frac{1}{98}-1-\frac{1}{97}+\frac{1}{97}-\frac{1}{98}\)
\(=1+1-1\)
\(=1\)
\(\frac{99}{98}-\frac{98}{97}+\frac{1}{97\times98}\)
\(=\left(1+\frac{1}{98}\right)-\left(1+\frac{1}{97}\right)+\frac{1}{97\times98}\)
\(=\frac{1}{98}-\frac{1}{97}+\frac{1}{97\times98}\)
\(=\frac{-1}{97\times98}+\frac{1}{97\times98}\)
\(=0\)
\(=\frac{96}{97}\cdot\frac{97}{98}\cdot...\cdot\frac{1997}{1998}=\frac{96\cdot97\cdot...\cdot1997}{97\cdot98\cdot...\cdot1998}=\frac{96\cdot1\cdot...\cdot1}{1\cdot1\cdot...\cdot1\cdot1998}=\frac{96}{1998}=\frac{16}{333}\)