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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
a) x + 5,84 = 9,16
=> x = 9,16 - 5,84
x = 3,32
b) \(x=\frac{1}{5}-\frac{3}{7}\)
=> x = \(\frac{-8}{35}\)
c) \(\frac{x}{5}=\frac{5}{6}-\frac{13}{30}\)
=> \(\frac{x}{5}=\frac{2}{5}\)
=> \(x=5\)
d) \(x+\frac{3}{4}=\frac{4}{5}\)
=> x = \(\frac{4}{5}-\frac{3}{4}\)
x = \(\frac{1}{20}\)
e) \(\frac{x}{7}=\frac{7}{8}-\frac{25}{56}\)
=> \(\frac{x}{7}=\frac{3}{7}\)
=> x = 3
t i c k nhé!! 426356457467568768467356457568578769578956
Câu a,x=9,16-5,84=3,32
Câu b, d bạn làm tương tự như câu a
Câu c,e bạn làm thế này, mik chỉ làm mẫu 1 câu thôi
c, x/5 = 25/30 - 13/30
x/5 = 12/30
x/5 = 2/5
x = 2/5 : 5
x = 2/5 X 1/5
x = 2/25
Vậy x=2/25
Câu e làm tương tự như vậy nhé.Bạn tính phần hiệu bên phải rồi lấy nó nhân với mẫu số của x là ra thôi.
Ủng hộ nhé
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
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
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
a)\(\frac{27}{5}\text{x}\frac{91}{12}\text{x}\frac{125}{9}\text{x}\frac{96}{13}=\frac{27}{9}\text{x}\frac{91}{13}\text{x}\frac{125}{5}\text{x}\frac{96}{12}=3\text{x}7\text{x}25\text{x}8\)
Đến đây tự tính nha !
b)\(\frac{2539}{35}\text{x}\frac{7}{90}+\frac{561}{35}\text{x}\frac{7}{90}=\frac{7}{90}\text{x}\left(\frac{2539}{35}+\frac{561}{35}\right)=\frac{7}{90}\text{x}\frac{3100}{35}=\frac{3100}{90}\text{x}\frac{7}{35}=\frac{310}{9}\text{x}\frac{1}{5}=\frac{62}{9}\)
`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`
\(M=3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{27}{12}\times3\dfrac{4}{7}-\dfrac{7}{25}\)
\(=3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{27}{12}\times\dfrac{25}{7}-\dfrac{7}{25}\)
\(=3+\left(\dfrac{4}{9}\times\dfrac{27}{12}\right)\times\left(\dfrac{7}{25}\times\dfrac{25}{7}\right)-\dfrac{7}{25}\)
\(=3+\left(\dfrac{4\times3\times9}{9\times3\times4}\right)\times1-\dfrac{7}{25}\)
\(=3+1\times1-\dfrac{7}{25}\)
\(=3+1-\dfrac{7}{25}\)
\(=4-\dfrac{7}{25}\)
\(=\dfrac{100}{25}-\dfrac{7}{25}\)
\(=\dfrac{93}{25}\)