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a: A=(-1+2)+(-3+4)+...+(-499+500)
=1+1+...+1=250
b: B=(2+4-6-8)+(10+12-14-16)+...+(394+396-398-400)
=(-8)x100=-800
c: \(C=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(997+998-999-1000\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot250=-1000\)
A=(1-2)+(3-4)+...+(999-1000)
có 1000 số hạng
A=(-1)+(*1)+...+(-1)
có 500 số hạng
A=-1*500
A=-500
a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000
=(1-2-3+4)+(5-6-7+8)+...+(1997-1998-1999+2000)
=0
b) B=1-3+5-7+....+2001-2003+2005
=(1-3)+(5-7)+...+(2001-2003)+2005
=-2.501+2005
=-1002+2005
=1003
c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997
=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997
=1997
d) D=1000+998+996+......+10-999-997-995-...-11
=(1000-999)+(998-997)+(996-995)+....+(12-11)+10
=1.495+10
=595
a) \(\frac{2}{5}+x=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{2}{5}\)
\(x=\frac{15}{20}-\frac{8}{20}\)
\(x=\frac{7}{20}\)
\(\)b)
\(x-\frac{1}{15}=\frac{3}{10}\\ x=\frac{3}{10}+\frac{1}{15}\\ x=\frac{9}{30}+\frac{2}{30}\\ x=\frac{11}{30}\)
c)
\(\frac{9}{8}-x=\frac{5}{12}\\ x=\frac{9}{8}-\frac{5}{12}\\ x=\frac{27}{24}-\frac{10}{24}\\ x=\frac{17}{24}\)
d)
\(\frac{3}{5}+x=\frac{5}{4}+\frac{7}{10}\\ \frac{3}{5}+x=\frac{25}{20}+\frac{14}{20}\\\frac{3}{5}+x=\frac{39}{20}\\ x=\frac{39}{20}-\frac{3}{5}\\ x=\frac{39}{20}-\frac{12}{20}\\ x=\frac{27}{20} \)
e)
\(\frac{9}{8}-x=\frac{3}{20}+\frac{2}{5}\\ \frac{9}{8}-x=\frac{3}{20}+\frac{8}{20}\\ \frac{9}{8}-x=\frac{11}{20}\\ x=\frac{9}{8}-\frac{11}{20}\\ x=\frac{45}{40}-\frac{22}{40}\\ x=\frac{23}{40}\)
g)
\(x+\frac{1}{3}=\frac{5}{6}+1\frac{7}{10}\\ x+\frac{1}{3}=\frac{5}{6}+\frac{17}{10}\\ x+\frac{1}{3}=\frac{25}{30}+\frac{51}{30}\\ x+\frac{1}{3}=\frac{76}{30}=\frac{38}{15}\\ x=\frac{38}{15}-\frac{1}{3}\\ x=\frac{38}{15}-\frac{5}{15}\\ x=\frac{33}{15}=\frac{11}{5}\)
h)
\(3x-\frac{3}{5}=\frac{1}{2}\\ 3x=\frac{1}{2}+\frac{3}{5}\\ 3x=\frac{5}{10}+\frac{6}{10}\\ 3x=\frac{11}{10}\\ x=\frac{11}{10}:3\\ x=\frac{11}{10}\cdot\frac{1}{3}\\ x=\frac{11}{30}\)
i)
\(4x+\frac{5}{12}+\frac{4}{9}=1\frac{13}{18}\\ 4x+\frac{5}{12}+\frac{4}{9}=\frac{31}{18}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{4}{9}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{8}{18}\\ 4x+\frac{5}{12}=\frac{23}{18}\\ 4x=\frac{23}{18}-\frac{5}{12}\\ 4x=\frac{46}{36}-\frac{15}{36}\\ 4x=\frac{31}{36}\\ x=\frac{31}{36}:4\\ x=\frac{31}{36}\cdot\frac{1}{4}\\ x=\frac{31}{144}\)
k)
\(2-\left(3x+\frac{3}{7}\right)=\frac{9}{21}\\ 2-\left(3x+\frac{3}{7}\right)=\frac{3}{7}\\3x+\frac{3}{7}=2-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{14}{7}-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{11}{7}\\ 3x=\frac{11}{7}-\frac{3}{7}\\ 3x=\frac{8}{7}\\ x=\frac{8}{7}:3\\ x=\frac{8}{7}\cdot\frac{1}{3}\\ x=\frac{8}{21} \)
A = -1+2-3+4-5+...+500
A = -1 +(2-3)+(4-5)+...+(498-499)+500
A = -1 + (-1) + (-1)+ ...+(-1) + 500 (có 250 số hạng -1)
A = -250 + 500 = 250
B = 2+4-6-8 + 10+12-...-398-400
B = (2+4-6-8)+(10+12-14-16)+...+(394+396-398-400)
B = -8 + (-8)+...+(-8) (có 50 số hạng -8)
B = -400
C = 1+2-3-4+5+6-7-8+...-999-100
C = (1+5+9+...+997)+[(2-3-4)+(6-7-8)+...+(998-999-100)]
C = (997+1).[(997-1)/4+1):2 + [(-5)+(-9)+...+(-1001)]
C = 124750 + -125750
C = -10