tính tổng S=\(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)
Tính hợp lý : \(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))
M= 3.(\(\frac{1}{5}-\frac{1}{61}\))
M=\(\frac{168}{305}\)
\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(M=\frac{84}{305}\)
Tính\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
Đặt A=như đã cho.
=>1/2A=2/5*7+2/7*9+2/9*11+...+2/59*61.
=>1/2A=1/5-1/7+1/7-1/9+1/9-1/11+...+1/59-1/61.
=>1/2A=1/5-1/61=56/305.
=>A=56/305*2=112/305.
k nha đúng đó.Có j kb nha.
bài 1 tính
A = \(\frac{2}{3.5}+\frac{2}{5.7}+..............+\frac{2}{37.39}\); B = \(\frac{4}{5.7}+\frac{4}{7.9}+..........+\frac{4}{59.61}\) ; C = \(\frac{4}{5.9}+\frac{4}{9.13}+.................+\frac{4}{41.45}\) Bài 2 chứng minh : \(\frac{m}{b.\left(b+m\right)}=\frac{1}{b}-\frac{1}{b+m}\)
\(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\) tìm A
\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{59}-\frac{1}{61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(\Rightarrow A=\frac{3}{2}.\frac{56}{305}\)
\(\Rightarrow A=\frac{84}{305}\)
\(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}?\)
=> A= \(\frac{3}{2}\) .( \(\frac{1}{5}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - \(\frac{1}{9}\) +...+ \(\frac{1}{59}\) - \(\frac{1}{61}\))
=> A=\(\frac{3}{2}\) . (\(\frac{1}{5}\) - \(\frac{1}{61}\) ) => A= \(\frac{3}{2}\). \(\frac{56}{305}\) = \(\frac{84}{305}\) Vậy A= \(\frac{84}{305}\)
\(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
GIẢI GIÚP MÌNH NHA
biểu thức trên =\(\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{56}-\frac{1}{61}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{1}{2}x\frac{61}{305}=\frac{1}{10}=0,1.\)
vậy biểu thức trên =0,1
\(B=\frac{4}{5.7}+\frac{4}{7.9}+......+\frac{4}{59.61}\) = ?
Ta có 1/2B=2/5.7+2/7.9+...+2/59.61
1/2B=1/5-1/7+1/7-1/9+1/9-...+1/59-1/61
1/2B=1/5-1/61
1/2B=56/305
B=56/305:1/2
B=112/305
A=\(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{59.61}\)=?
\(\frac{2}{3}A=\frac{2}{3}.\left(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\right)\)
\(\frac{2}{3}A=\frac{2.3}{3.5.7}+\frac{2.3}{3.7.9}+...+\frac{2.3}{3.59.61}\)
\(\frac{2}{3}A=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)
\(\frac{2}{3}A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(\frac{2}{3}A=\frac{1}{5}-\frac{1}{61}\)
\(\frac{2}{3}A=\frac{56}{305}\)
\(A=\frac{56}{305}.\frac{3}{2}\)
\(A=\frac{84}{305}\)
\(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{3}{59.61}\)
\(A=\frac{1}{5}-\frac{1}{61}\)
\(A=\frac{56}{305}\)
Help me do my homework
\(\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
= 2/2 . ( 4 / 5.7 +4 / 7.9 +...+ 4 / 59.61 )
= 4/2 . ( 2 / 5.7 +2 / 7.9 +...+ 2 / 59.61 )
= 2 . ( 7-5 / 5.7 + 9-7 / 7.9 +...+ 61-59 / 59.61 )
= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 +...+ 1/59 - 1/61 )
= 2 . ( 1/5 - 1/61 )
= 2 . ( 61/305 - 5/305 )
= 2 . 56/305
= 112/305
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