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Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
\(G=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
\(G=2.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{504.509}\right)\)
\(G=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(G=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(G=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(G=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
1) -2/2.3+(-2/3.4)+(-2/4.5)+...+(-2/19.20)
=-1(1/2-1/3+1/3-1/4+...+1/19-1/20)
=-1(1/2-1/20)
=-1.9/20
=-9/20
à nhầm
1)=-2(1/2-1/3+1/3-1/2+...+1/19-1/20)
=-2.(1/2-1/20)
=-2.9/20
=-9/10
Tìm x,y thuộc N biết:
a/ ( 3.x-2).(2y-3)=1
b/ (x-5).(x+1)=7
c/ (x+1)>(2y-1)=10
giải giúp mik bài này nhé
a)(3x-2)(2y-3)=1
Ta xét bảng sau:
3x-2 | 1 |
3x | 3 |
x | 1 |
2y-3 | 1 |
2y | 4 |
y | 2 |
b)(x-5)(x+1)=7
Ta xét bảng sau:
x-5 | 1 | 7 |
x | 6 | 12 |
x+1 | 7 | 1 |
x | 6 | 0 |
=>x=6
c)mk chả hiểu cậu ghi j hết
a)
\(6x+5=4x-7\)
\(\Leftrightarrow6x-4x=-7-5\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
b)
\(-3\left(x-5\right)-1=2x-1\)
\(\Leftrightarrow-3x+15-1=2x-1\)
\(\Leftrightarrow-3x-2x=-1-15+1\)
\(\Leftrightarrow-5x=-15\)
\(\Leftrightarrow x=3\)
a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)
\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)
\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)
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