Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=-[-506+732-(-2000)]-(506-1732)
A=-[226+2000]-506+1732
A=1774-506+1732
A=1268+1732
A=3000
K MK NHE
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
Ta có : [ 210 + ( 59 - x ) ] : 2 + 14 = 144
=> [ 210 + ( 59 - x ) ] : 2 = 130
=> 210 + ( 59 - x ) = 260
=> 59 - x = 50
=> x = 9
Vậy x = 9 là giá trị cần tìm
a = 2002.2002 = 2002 . (2000+2) = 2002.2000 + 2002.2
b = 2000.2004 = 2000 . (2002+2) = 2000.2002 + 2000.2
Vậy a>b.
a > b nhé vì :
a = 2002.2002 = 2002 .( 2002 + 2 ) = 2002.2000 + 2002.2
b = 2000.2004 = 2000 .(2002 + 2 ) = 2000.2002 + 2000.2
nên a > b
\(a,\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)
\(=\left(\dfrac{-3}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}+-\dfrac{4}{5}\)
\(=\dfrac{4}{5}+\dfrac{7}{21}+-\dfrac{4}{5}\)
\(=\left(\dfrac{4}{5}+\dfrac{-4}{5}\right)+\dfrac{7}{21}\)
\(=0+\dfrac{7}{21}=\dfrac{7}{21}\)
\(b,\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
\(=-\dfrac{3}{17}+\dfrac{2}{3}+\dfrac{3}{17}\)
\(=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}=\dfrac{2}{3}\)
\(c,\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\dfrac{-5}{21}+\dfrac{-16}{21}+1\)
\(=\dfrac{-21}{21}+1\)
\(=\left(-1\right)+1=0\)
Tick mình nha ^^
a.\(\dfrac{-3}{5}\)+\(\dfrac{7}{21}\)+\(\dfrac{-4}{5}\)+\(\dfrac{7}{5}\)= (\(\dfrac{-3}{5}\)+\(\dfrac{-4}{5}\)+\(\dfrac{7}{5}\)) + \(\dfrac{7}{21}\)=0+\(\dfrac{7}{21}\)=\(\dfrac{7}{21}\)
b.\(\dfrac{-3}{17}\)+(\(\dfrac{2}{3}\)+\(\dfrac{3}{17}\))= \(\dfrac{-3}{17}\)+\(\dfrac{3}{17}\)+\(\dfrac{2}{3}\)=0+\(\dfrac{2}{3}\)=\(\dfrac{2}{3}\)
c.\(\dfrac{-5}{21}\)+(\(\dfrac{-16}{21}\)+1)=\(\dfrac{-5}{21}\)+\(\dfrac{-16}{21}\)+1=(-1)+ 1 = 0