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đối với câu a thì bạn phân tích ra nha:
ta có:
A = \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-8}{10^{2005}}+\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}\)
vì \(\frac{8}{10^{2005}}>\frac{8}{10^{2006}}=>\frac{-8}{10^{2005}}< \frac{-8}{10^{2006}}\)
=> A > B
CÂU b mk làm phân số hơi mất thời gian nên bn thông cảm cho mk nha:
1/5*8 + 1/8*11 + 1/11*14 +...+ 1/x(x+3) = 101/1540
=> 1/5 - 1/8 + 1/8 - 1/11 + 1/11 -...+ (1/x) - (1/ x+3) = 101/1540
=>1/5 - 1/x+3 = 101/1540
=> 1/x+3 = 1/5 - 101/1540
=> 1/x+3 = 1/308
=> 308*1 = (x+3)*1
=> 308 = x+3
=> x = 308 - 3
=> x = 305
Chúc bn học tốt !
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+............+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.............-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{49}{98}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{48}{98}\)
\(A=\frac{8}{49}\)
Vậy A = \(\frac{8}{49}\)
Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 = 1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)
A = \(\frac{1}{2}-\frac{1}{98}\)
A = \(\frac{24}{49}\)
Vậy A = \(\frac{24}{49}\)
~~~
#Sunrise
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)
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 !!
\(A=\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+\frac{4}{11\cdot14}+...+\frac{4}{305\cdot308}\)
\(\frac{3}{4}A=\frac{3}{4}\left(\frac{4}{5\cdot8}+\frac{4}{8\cdot11}+...+\frac{4}{305\cdot308}\right)\)
\(\frac{3}{4}A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{305\cdot308}\)
\(\frac{3}{4}A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\)
\(\frac{3}{4}A=\frac{303}{1540}\Rightarrow A=\frac{303}{1540}:\frac{3}{4}=\frac{101}{385}\)
A=1/3x(1/2x5+1/5x8+......+1/95x98)
A=1/3x(1/2-1/5+1/5-1/8+.........+1/95-1/98)
A=1/3x(1/2-1/98)
A=1/3x24/49
A=8/49
A =\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1.3}{2.5.3}+\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+...+\frac{1.3}{92.95.3}+\frac{1.3}{95.98.3}\)
A = \(\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
A =\(\frac{1}{3}.\frac{97}{98}\)
A =\(\frac{97}{294}\)
- a) giai ta co (1/2.5=1/2-1/5)+(1/5.8=1/5-1/8)+....+(1/2009.2012=1/2009-1/2012) =>a=1/2-1/5+1/5-1/8+...+1/2009-1/2012 <=>1/2-1/2012 =>a=1005/2012 câu b bằng nhau nhhung minh không th
- e giải ra được
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{21.24}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{24}\)
\(=\frac{1}{5}-\frac{1}{24}\)
\(=\frac{19}{120}\)
Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(A=\frac{1}{2}-\frac{1}{17}\)
\(A=\frac{15}{34}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)
_Sai đề
Cái phân số cuối cùng phải là 15/92.95