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\(=\dfrac{1\cdot2\cdot3+8\cdot1\cdot2\cdot3+64\cdot1\cdot2\cdot3+125\cdot1\cdot2\cdot3}{1\cdot3\cdot4+8\cdot1\cdot3\cdot4+64\cdot1\cdot3\cdot4+125\cdot1\cdot3\cdot4}\)
\(=\dfrac{1\cdot2\cdot3}{1\cdot3\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)
Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+......+\frac{2}{19\times20}\)
\(=2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{19\times20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}\)
3 \(\times\) \(\dfrac{4}{15}\) + 2 \(\times\) \(\dfrac{4}{15}\) - 5 \(\times\) \(\dfrac{4}{15}\)
= \(\dfrac{4}{15}\) \(\times\) ( 3 + 2 - 5)
= \(\dfrac{4}{15}\) \(\times\) 0
= 0
a: \(=\dfrac{3\cdot\left(6+4\right)}{6\cdot2}=\dfrac{3\cdot10}{6\cdot2}=\dfrac{30}{12}=\dfrac{5}{2}\)
b: \(=\dfrac{12\left(6-1\right)}{3\left(4+5\right)}=4\cdot\dfrac{5}{9}=\dfrac{20}{9}\)
1/1x2+1/2x3+1/3x4+..+1/9x10
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...-1/10
=1-1/10
=9/10
a) Ta có \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
= \(\frac{1.2+1.2.2.2+1.3.2.3+1.4.2.4+1.5.2.5}{3.4+3.2.4.2+3.3.4.3+3.4.4.4+3.5.4.5}\)
= \(\frac{1.2+1.2.4+1.2.9+1.2.16+1.2.25}{3.4+3.4.4+3.4.9+3.4.16+3.4.25}\)
= \(\frac{1.2.\left(1+4+9+16+25\right)}{3.4.\left(1+4+9+16+25\right)}\)
= \(\frac{1.2.55}{3.4.55}\)
= \(\frac{1.2.55}{3.2.2.55}\)
= \(\frac{1}{3.2}\)
= \(\frac{1}{6}\)
b) \(\frac{111111}{666665}=\frac{111111}{666665}\)
(dấu "." là dấu "x")
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\)\(1-\frac{1}{2014}\)
\(=\)\(\frac{2014}{2014}-\frac{1}{2014}\)
\(=\)\(\frac{2013}{2014}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{2013}{2014}\)
Dấu \(.\) là dấu nhân nhé
Chúc bạn học tốt ~
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2013\times2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)
CHÚC BN HỌC TỐT!!!!!
x là nhân
Trả lời:
\(\frac{2\times4+2\times4\times8+4\times8\times16+8\times16\times32}{3\times4+2\times6\times8+4\times12\times16+8\times24\times32}\)
\(=\frac{2\times4+2\times4\times8+2.2\times2.4\times16+4.2\times4.4\times32}{3\times4+2\times2.3\times2.4+4\times3.4\times4.4+8\times8.3\times8.4}\)
\(=\frac{2\times4\times\left(1+8+2\times2\times16+4\times4\times32\right)}{3\times4\times\left(1+2\times2\times2+4\times4\times4+8\times8\times8\right)}\)
\(=\frac{2\times4\times\left(1+8+64+512\right)}{3\times4\times\left(1+8+64+512\right)}\)
\(=\frac{2}{3}\)