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Ta có:
\(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)
=\(\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{-1}{8}+\frac{1}{8}\right)+\frac{1}{9}\)
=\(\left(\frac{-1+1}{5}\right)+\left(\frac{-1+1}{6}\right)+\left(\frac{-1+1}{7}\right)+\left(\frac{-1+1}{8}\right)+\frac{1}{9}\)
=\(\frac{0}{5}+\frac{0}{6}+\frac{0}{7}+\frac{0}{8}+\frac{1}{9}\)
=\(\frac{1}{9}\)
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: \(\dfrac{1}{5}+\dfrac{-1}{6}+\dfrac{1}{7}+\dfrac{-1}{8}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{-1}{7}+\dfrac{1}{6}+\dfrac{-1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(-\dfrac{1}{8}+\dfrac{1}{8}\right)+\dfrac{1}{9}\)
\(=0+0+0+0+\dfrac{1}{9}=\dfrac{1}{9}\)
5: \(=\dfrac{43}{8}-\dfrac{19}{10}=\dfrac{430}{80}-\dfrac{152}{80}=\dfrac{278}{80}=\dfrac{139}{40}\)
6: \(=\dfrac{-13}{4}-\dfrac{7}{3}=\dfrac{-39}{12}-\dfrac{28}{12}=\dfrac{-67}{12}\)
7: \(=\dfrac{-41}{8}+\dfrac{14}{4}=\dfrac{-41}{8}+\dfrac{28}{8}=\dfrac{-13}{8}\)
8: \(=\dfrac{-43}{7}+\dfrac{-43}{6}=\dfrac{-559}{42}\)
5, \(5\dfrac{3}{8}-1\dfrac{9}{10}=\dfrac{139}{40}\)
6, \(\left(-3\dfrac{1}{4}\right)+\left(-2\dfrac{1}{3}\right)=-\dfrac{67}{12}\)
7, \(\left(-5\dfrac{1}{8}\right)+3\dfrac{2}{4}=-\dfrac{13}{8}\)
8, \(\left(-6\dfrac{1}{7}\right)+\left(-7\dfrac{1}{6}\right)=-\dfrac{559}{42}\)
Ta có:
1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
=1-1/10
=9/10
( 1+2+3+4-1-2-3-4) * 1*2*...........*8*9
= 0 * 1*2*...........*8*9
= 0
(1.2.3.4.5.6.7.8.9) - (1.2.3.4.5.6.7.8) - (1.2.3.4.5.6.7.82)
= (1.2.3.4.5.6.7.8)(9 - 1 - 8)
= (1.2.3.4.5.6.7.8).0
= 0
Chúc bn học tốt! (Ez :D)