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Bài 1 :
\(\frac{18}{23}:9=\frac{2}{23}\) \(8:\frac{3}{4}:\frac{4}{9}=\frac{32}{3}:\frac{4}{9}=24\)
Bài 2 :
0,25 x 6,9 x 4
= 0,25 x 4 x 6,9
= 1 x 6,9
= 6,9
7,2 x 4,7 + 72 : 0,1 + 7,2 x 4,3
= 7,2 x 4,7 + 7,2 x 10 + 7,2 x 4,3
= 7,2 x ( 4,7 + 10 + 4,3 )
= 7,2 x 19
= 136,8
14,7 x 31 + 29,4 + 67 x 14,7
= 14,7 x 31 + 14,7 x 2 + 67 x 14,7
= 14,7 x ( 31 + 2 + 67 )
= 14,7 x 100
= 1470
20,19 x 6 + 20,19 + 3 x 20,19
= 20,19 x ( 6 + 1 + 3 )
= 20,19 x 10
= 201,9
~ Thiên Mã ~
\(\frac{18}{23}.\frac{1}{9}=\frac{2}{23}.\frac{1}{1}=\frac{2}{23}\\ 8.\frac{4}{3}.\frac{9}{4}=8.\frac{1}{3}.\frac{9}{1}=\frac{8}{3}.9=\frac{72}{3}=4\\ mìnhchỉghicáchlàmkhôngghiđềbàinha\)
a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\)
\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)
\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)
b: =>2(x+7/8)=6*13/4=78/4=39/2
=>x+7/8=39/4
=>x=71/8
`3+4/9xx7/25xx27/12xx3 4/7-7/25`
`=3+4/9xx7/25xx27/12xx25/7-7/25`
`=3+7/25xx25/7xx4/9xx27/12-7/25`
`=3+4/9xx9/4xx1-7/25`
`=3+1xx1-7/25`
`=3+1-7/25`
`=75/25+25/25-7/25`
`=93/25`
\(M=3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{27}{12}\times3\dfrac{4}{7}-\dfrac{7}{25}\)
\(=3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{27}{12}\times\dfrac{25}{7}-\dfrac{7}{25}\)
\(=3+\left(\dfrac{4}{9}\times\dfrac{27}{12}\right)\times\left(\dfrac{7}{25}\times\dfrac{25}{7}\right)-\dfrac{7}{25}\)
\(=3+\left(\dfrac{4\times3\times9}{9\times3\times4}\right)\times1-\dfrac{7}{25}\)
\(=3+1\times1-\dfrac{7}{25}\)
\(=3+1-\dfrac{7}{25}\)
\(=4-\dfrac{7}{25}\)
\(=\dfrac{100}{25}-\dfrac{7}{25}\)
\(=\dfrac{93}{25}\)
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
Gợi ý :
a) tìm khoảng cách -> số số hạng -> tổng
b) và c) : đặt nhân tử chung rồi tính nốt phần còn lại
\(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)
\(=\left(\dfrac{1}{10}+\dfrac{9}{10}\right)+\left(\dfrac{2}{10}+\dfrac{8}{10}\right)+\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{4}{10}+\dfrac{6}{10}\right)+\dfrac{5}{10}\)
\(=1+1+1+1+\dfrac{5}{10}\)
\(=4+\dfrac{5}{10}\)
\(=\dfrac{45}{10}\)
\(13,25:0,5+13,25:0,25+13,25:0,125+13,25\times6\)
\(=13,25:\dfrac{1}{2}+13,25:\dfrac{1}{4}+13,25:\dfrac{1}{8}+13,25\times6\)
\(=13,25\times2+13,25\times4+13,25\times8+13,25\times6\)
\(=13,25\times\left(2+4+8+6\right)\)
\(=13,25\times20\)
\(=265\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a) 7,2 x 94,3 + 7,2 x 4,7 + 7,2
=7,2 x(94,3+4,7+1)
=7,2x100
=720
b)\(\dfrac{7}{9}\times\dfrac{2}{3}+\dfrac{7}{9}\times\dfrac{1}{3}=\dfrac{7}{9}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{7}{9}\times1=\dfrac{7}{9}\)