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20,03 x 7,5 + 200,3 x 0,25
= 150,225 + 50,075
= 200,3
11 x 7 - 700 x 0,1 - 7
= 77 - 70 - 7
= 7 - 7
= 0
k mk nha
Mk cảm ơn các bạn nhiều
Thank you very much
( ^ _ ^ )
a)20,03x7,5+20,03x2,5 b)11x7-7x10-7 c)40x9,84x0,25 d)46:24+8:24 e)705:45-336:45 f)23,45:12,5:0,8
=20,03x(7,5+2,5) =11x7-7x10-7x1 =9,84x(40x0,25) =(46+8):24 =(705-336):45 =1,876:0.8
=20,03x10 =7x(11-10-1) =9,84x10 =54:24 =369:45 =2,345
=200,3 =7x0 =98,4 =2,25 =8,2
=0
CHÚC BẠN HỌC TỐT !!!!!!!!!!!!
20,03 . 7,5 + 20,03 . 2,5 11 . 7 - 7 . 10 -7 40 . 9,84 . 0,25
= 20,03 . ( 7,5 + 2,5 ) = 7 . ( 11 - 10 ) = ( 40 . 0,25 ) . 9,84
= 20,03 . 10 = 7 .1 = 10 . 9,84
= 200,3 = 7 = 98 ,4
46 : 24 + 8 : 24 705 : 45 - 336 : 45
= ( 46 + 8 ) ; 24 = ( 705 - 336 ) : 45
= 54 : 24 = 369 : 45
= 2,25 = 8,2
Sửa đề: Tính M/N
\(\dfrac{M}{N}=\dfrac{\dfrac{7}{4}-\dfrac{7}{18}-\dfrac{7}{19}}{\dfrac{11}{4}-\dfrac{11}{18}-\dfrac{11}{19}}\)
\(=\dfrac{7\left(\dfrac{1}{4}-\dfrac{1}{18}-\dfrac{1}{19}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{18}-\dfrac{1}{19}\right)}=\dfrac{7}{11}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
\(\dfrac{1}{7}\) + \(\dfrac{3}{11}\) + \(\dfrac{2}{7}\) + \(\dfrac{8}{11}\) + \(\dfrac{4}{7}\) + 2010
= ( \(\dfrac{1}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{4}{7}\)) + ( \(\dfrac{3}{11}\) + \(\dfrac{8}{11}\)) + 2010
= \(\dfrac{7}{7}\) + \(\dfrac{11}{11}\) + 2010
= 1 + 1 + 2010
= 2012
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
Sửa đề:
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{28}{1995}\)
\(=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{285}\)
\(=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}\)
\(=\frac{1}{3}-\frac{1}{19}\)
\(=\frac{16}{57}\)