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=> D = 1/2-1/5+1/5-1/8+....+ 1/62-1/65
=> D= 1/2-1/65
=> D=63/130
VẬY D=63/130
\(D=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\cdot\cdot\cdot+\frac{3}{62\cdot65}\)
\(\Rightarrow D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\cdot\cdot\cdot+\frac{1}{62}-\frac{1}{65}\)
\(\Rightarrow D=\frac{1}{2}-\frac{1}{65}\)
\(\Rightarrow D=\frac{63}{130}\)
\(=\frac{3}{4}\cdot\left(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{62.65}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\)
\(=\frac{1}{2}-\frac{1}{65}\)
\(=\frac{63}{130}\)
Đặt A=4/2.5+4/5.8+4/8.11+...+4/62.65.Ta có A=4.(1/2.5+1/5.8+1/8.11+...1/62.65)=4/3.(3/2.5+3/5.8+3/8.11+...+3/62.65) =4/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+3/62-3/65)=4/3.(1/2-1/65)=4/3.63/130=42/56 Vậy A=42/56
a) e chỉ cần nhân chúng lại với nhau = cách tách từng cái ra
b)đặt 4/2.5+4/5.8+4/8.11+......+4/62.65 là S
\(.S=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\right)\)
\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(S=\frac{4}{3}\left(\frac{65}{130}-\frac{2}{130}\right)\)
\(S=\frac{4}{3}\left(\frac{63}{130}\right)\)
\(S=\frac{42}{65}\)
Bài 2 : \(\frac{15+a}{29+a}=\frac{3}{5}\)\(\Leftrightarrow\left(15+a\right)5=\left(29+a\right)3\Leftrightarrow75+5a=87+3a\Leftrightarrow5a-3a=87-75\Rightarrow2a=12\Rightarrow a=6\)
vậy a =6
\(\frac{3}{2.5}\)+ \(\frac{3}{5.8}\)+ ...... + \(\frac{3}{92.95}\)= 3 . ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ .... + \(\frac{1}{92.95}\))
= 3 . \(\frac{1}{3}\). ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ ..... + \(\frac{1}{92.95}\))
= 3. \(\frac{1}{3}\). ( \(\frac{1}{2}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{8}\)+ ....... + \(\frac{1}{92}\)- \(\frac{1}{95}\))
= 1 .( \(\frac{1}{2}\)- \(\frac{1}{95}\)) = \(\frac{93}{190}\)
Thấy hay thì cho mình một k nhé!!!
3/ 2.5 + 3/ 5.8 + 3/ 8.11+ ...+ 3/ 92.95
=1/2-1/5+1/5-1/8+1/8-1/11+........+1/92-1/95
=1/2-1/95
=31/60
3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
3/2.5 + ...+ 3/17 .20
= 3/2 .(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 120)
= 3/2 . (1/2 - 1/20)
= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{62.65}\)
\(=1.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(=1.\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(=1.\frac{63}{130}\)
\(=\frac{63}{130}\)
Bài làm
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{62.65}\)
\(=3.\frac{1}{2.5}+3.\frac{1}{5.8}+3.\frac{1}{8.11}+...+3.\frac{1}{62.65}\)
\(=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{62.65}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{62}-\frac{1}{65}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{65}\right)\)
\(=3.\left(\frac{65}{130}-\frac{2}{130}\right)\)
\(=3.\frac{63}{130}\)
\(=\frac{3.63}{130}\)
\(=\frac{189}{130}\)
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