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\(\frac{1}{10}+\frac{20}{100}+\frac{300}{1000}+\frac{40000}{100000}\)
\(=\frac{1}{10}+\frac{20}{10}+\frac{30}{10}+\frac{40}{10}\)
\(=\frac{1+2+3+4}{10}\)
\(=\frac{10}{10}\)
\(=1\)
1/10 + 20/100 + 300/1000 + 4000/10000
=1/10 + 2/10 + 3/10 + 4/10
=10/10
=1
1000:[50+50+(50x2+300)]
=1000:[50+50+(100+300)]
=1000:(50+50+400)
=1000:(100+400)
=1000:500
=2
a/ ( -500 ) + (-174) + 1999 + (-266) + ( - 1499 )
=[(-174)+(-266)]+[1999+(-1499)+(-500)]
=(-440)+0
=-440
b/ 1000 + ( - 670 ) + 2004 + ( -300 )
=2004+1000-970
=2004+30
=2034
Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
1 - 1000 + 212 - 300 = -1087.
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Ai tk cho mình mình tk lại.
1-1000+212-300=
=-999+212-300
=-787-300
=-1087