tính giá trị của biểu thức:
\(P=\frac{15}{5.8}+\frac{15}{8.11}+\frac{15}{11.14}+...+\frac{15}{92.95}\)
\(\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-...-\frac{15}{47.50}\)
Ta có : \(\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-......-\frac{15}{47.45}\)
\(=\frac{3}{8}-\left(\frac{15}{8.11}+\frac{15}{11.14}+\frac{15}{14.17}+......+\frac{15}{47.50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{11}{47}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\frac{1}{8}+\frac{1}{50}\)
\(=\frac{1}{4}+\frac{1}{50}=\frac{27}{100}\)
\(3x-\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-...-\frac{15}{47.50}=2\frac{1}{10}\)
\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)
<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)
<=> \(3x-\frac{9}{10}=\frac{21}{10}\)
<=> \(3x=3\)
<=> \(x=1\)
a.So sánh (không quy đồng): \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}};B=\frac{-15}{10^{2005}}+\frac{_{-7}}{10^{2006}}\)
b.Tìm x: \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
Làm đc mình tick cho!!!!!!
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 !
b.Đặt S=\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\) ta có
\(S=\frac{1}{3}.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
Tính:
\(\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2003}\right).\left(-1\frac{1}{2004}\right)\)
A=\(\frac{15}{2.5}+\frac{15}{5.8}+\frac{15}{8.11}....+\frac{15}{32.35}\)
Tính giá trị của các biểu thức sau:
a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}}\); b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8}\).
a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)
b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)
Tính
A) \(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+...+\(\frac{1}{605.608}\)
B) A=\(\frac{3}{2^2}\).\(\frac{8}{3^2}\).\(\frac{15}{4^2}\)....\(\frac{899}{30^2}\)
TÍNH
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)
\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\cdot\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
\(1-\frac{1}{2.5}-\frac{1}{5.8}-..-\frac{1}{92.95}=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}=1-\frac{31}{190}=\frac{159}{190}\)
học tốt nha
Tính: A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
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=1/3.(1/2-1/5 + 1/5 - 1/8 +......+1/92 - 1/95 + 1/95 - 1/98)
A=1/3.(1/2 - 1/98)
A=1/3. 48/98
A=48/294
Theo mk thì như vậy
Chúc bạn hok tốt ^O^
Tính giá trị của biểu thức:
A = \(\frac{38}{5}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-....+\frac{197}{4851}-\frac{199}{4950}\)