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c) =-5/7x3/9+4/9x3/-7
=-5/7x3/9+3/9x4/-7
=3/9x(-5/7+4/-7)
=3/9x-9/7
=-3/7
d) =58/7 - (31/9+30/7)
=58/7-31/9-30/7
=(58/7-30/7)-(31/9)
=4-31/9
=5/9
A = 4/5.8 + 4/8.11 + ... + 4/305.308
A = 4. ( 1/5.8 + 1/8.11 + ... + 1/305 . 108 )
A = 4 x 1/3 ( 3/5.8 + 3/8.11 + ... + 3/105.108 )
A = 4/3 ( 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/105 - 1/108 )
A = 4/3 ( 1/5 - 1/108 )
A = 4/3 . 103/540
A = 617/540
Vậy A = 617/540 !!
\(A=\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+\frac{4}{11\cdot14}+...+\frac{4}{305\cdot308}\)
\(\frac{3}{4}A=\frac{3}{4}\left(\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+...+\frac{4}{305\cdot308}\right)\)
\(\frac{3}{4}A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{305\cdot308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{303}{1540}\Rightarrow A=\frac{303}{1540}:\frac{3}{4}=\frac{101}{385}\)
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)
4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)
4\(A=\frac{1}{6}-\frac{1}{406}\)
4\(A=\frac{100}{609}\)
\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)
=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406
=1/6-1/406