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\(S=2.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=1-\frac{1}{16}< 1\)
Vậy \(S< 1\)
1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)
c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)
từ rồi làm tiếp
Bài 5 :
S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399
S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)
S = 1 + 0 + 0 + ... + 0 + (- 401)
S = 1 - 401
S = - 400
Bài 5
A= 1+3-5-7+9+11-13-15+...-397-399
A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)
A= -8 -8 -...-8
A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)
A= -400
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)
`Answer:`
Vì đề bài khá khó nhìn nên mình làm thành hai bài nhé. Đề nào đúng thì bạn tham khảo.
`x : (9 1/2 - 3/2 ) = 0,4 + 2/9 - 2/11 . 1/6 + 8/9 - 8/11`
`=> x : ( 19/2 - 3/2 ) = 28/45 - 1/33 + 8/9 - 8/11`
`=> x : 8 = 293/495 + 8/9 - 8/11`
`=> x : 8 = 733/495 - 8/11`
`=> x : 8 = 373/495`
`=> x = 373/495 .8`
`=> x = 2984/495`
`x : ( 9 1/2 - 3/2 ) = (0,4 + 2/9 - 2/11 )/(1,6 + 8/9 - 8/11 )`
`=> x : (19/2 - 3/2 ) = (2 . (1/5 + 1/9 - 1/11))/(8.(1/5 + 1/9 - 1/11))`
`=> x : 8 = 2/8`
`=> x = 2/8 .8`
`=> x = 2`
2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
Bài 1 :
S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)
= 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....