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2323/202+2323/606+2323/1212+2323/2020+2323/3030+2323/4242+2323/5656+2323/7272+2323/9090.=
\(\frac{23\times101}{2\times101}+\frac{23\times101}{6\times101}+\frac{23\times101}{12\times101}+....+\frac{23\times101}{72\times101}+\frac{23\times101}{90\times101} \)
=\(\frac{23}{2}+\frac{23}{6}+..........+\frac{23}{72}+\frac{23}{90}\)
=\(23\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+.....+\frac{1}{9\times10}\right)\)
=\(23\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}_{ }\right)\)
=\(23\left(1-\frac{1}{10}\right)_{ }\)
=\(23\times\frac{9}{10}\)
==\(\frac{207}{10}\)
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Gọi \(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\) là A
Ta sẽ tính biểu thức A.\(\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=A.\(\left(\dfrac{7}{12}-\dfrac{7}{12}\right)=0\)
Vậy \(\left(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\right).\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=0
\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)
\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)
\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)
\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)
\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)
\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)
\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\) ( 4^3 = 8^2 ; 5^4 = 25^2 )
\(=\frac{1}{25\times7}=\frac{1}{175}\)
\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)
\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)
\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)
\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)
a)Ta có : \(\dfrac{1212}{1414}=\dfrac{12.101}{14.101}=\dfrac{12}{14}\)
\(\dfrac{121212}{242424}=\dfrac{12.10101}{24.10101}=\dfrac{12}{24}\)
Vậy \(\dfrac{12}{24}=\dfrac{1212}{2424}=\dfrac{121212}{242424}\)
Ta có : \(\dfrac{2424}{3535}=\dfrac{24.101}{35.101}=\dfrac{24}{35}\)
\(\dfrac{242424}{353535}=\dfrac{24.10101}{35.10101}=\dfrac{24}{35}\)
Vậy \(\dfrac{24}{35}=\dfrac{2424}{3535}=\dfrac{242424}{353535}\)
=\(70\left(\frac{121212}{565656}+\frac{121212}{727272}+\frac{121212}{909090}\right)\)
=\(70\left(\frac{3.40404}{14.40404}+\frac{121212}{6.121212}+\frac{2.60606}{15.60606}\right)\)
=\(70\left(\frac{3}{14}+\frac{1}{6}+\frac{2}{15}\right)\)
=\(70.\frac{18}{35}=38\)
\(\dfrac{121212}{353535}+-\dfrac{2323}{4242}\)
\(=\dfrac{12}{35}+-\dfrac{23}{42}\)
\(=-\dfrac{43}{210}\)
\(=\dfrac{12}{35}-\dfrac{23}{42}=\dfrac{72-115}{210}=\dfrac{-43}{210}\)