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a: \(\dfrac{3}{5}-\left(-\dfrac{1}{2}\right)=\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{6+5}{10}=\dfrac{11}{10}\)
b: \(-\dfrac{48}{96}+\dfrac{37}{148}\)
\(=-\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=-\dfrac{1}{4}\)
c: \(\dfrac{1}{5}+\dfrac{-1}{6}+\dfrac{1}{7}+\dfrac{-1}{8}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{-1}{7}+\dfrac{1}{6}+\dfrac{-1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(-\dfrac{1}{8}+\dfrac{1}{8}\right)+\dfrac{1}{9}\)
\(=0+0+0+0+\dfrac{1}{9}=\dfrac{1}{9}\)
A = 100 + 98 + 96 + ... + 2 - 97 - 95 - 93 - ... - 1
A = (100 + 98 + 96 + ... + 2) - (97 + 95 + 93 + ... + 1)
A = 2550 - 2401
A = 149
a/ A= 1-3+5-7+9-11+......+97-99
= -2+(-2)+(-2)+......+(-2)
= (-2).25=-50
b/B=-1-2-3-4-...-100
=-(1+2+3+4+...+100)
=-5050
c/C=1-2+3-4+5-6+......+99-100
= -1+(-1)+(-1)+.............+(-1)
=(-1).50=-50
d/D=1-2-3+4+5-6-7+8+9-....+94-95
= (1-2-3+4)+(5-6-7+8)+.......+(92-93-94+95)
= 0+0+0+...+0=0
\(A=100+98+96+...+2-97-95-...-1\)
\(A=100+\left(98-98\right)+\left(96-95\right)+...+\left(2-1\right)\)
\(A=100+1+1+...+1\)
\(A=100+1\cdot49\)
\(A=100\cdot49\)
\(A=4900\)
\(B=1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302\)
\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(298-299-300+301\right)+302\)
\(B=1+0+0+...+302\)
\(B=1+302\)
\(B=303\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(2.A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(2.A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(2.A=\frac{1}{3}-\frac{1}{39}\)
\(2.A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
\(A=\frac{4}{13}:2=\frac{4}{13}.\frac{1}{2}=\frac{2}{13}\)
\(B=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{95.96}\)
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{95}-\frac{1}{96}\)
\(B=\frac{1}{3}-\frac{1}{96}\)
\(B=\frac{32}{96}-\frac{1}{96}=\frac{31}{96}\)