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`x+0,25=18/5+43/4`
`x+0,25=3,6+10,75`
`x=3,6+10,5`
`x=14,1`
Vậy `x=14,1`
`x+0,25=18/5+43/4`
`⇔x+1/4=18/5+43/4`
`⇔ x-18/5=43/4-1/4`
`⇔ x-18/5=42/4=21/2`
`⇔ x-36/10=105/10`
`⇔ x=141/10`
Phần 2: 1998 x 1996 + 1997 x 11 + 1985 / 1997 x 1996 - 1995 x 1996
=1998 x 1996+ 1997 x 11 +1985/1997x1996 - 1995 x 1996
=1998x11+1985/1995x1996
=23963/3982020
không chắc lắm đâu,câu này bạn hỏi rồi mà
a) \(x+\frac{18}{5}=\frac{43}{4}\)
\(x=\frac{43}{4}-\frac{18}{5}\)
\(x=\frac{143}{20}=7,15\)
b) \(x-4,2=42,19\)
\(x=42,19+4,2\)
\(x=46,39\)
Ta nhận xét quan hệ giữa hai phân số \(\frac{767}{767}\&\frac{767767}{767767}\)ta thấy \(\frac{767}{767}\times1001=\frac{767767}{767767}\)nên hai phân số này bằng nhau.
Ta nhận xét quan hệ giữa hai phân số \(\frac{15}{26}\&\frac{17}{13}\)ta thấy \(\frac{15}{26}<1\) , \(\frac{17}{13}>1\)nên \(\frac{15}{26}<\frac{17}{13}\)
\(\)
=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 = \(\dfrac{502}{3}\)
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
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; (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
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 ~
\(\frac{18}{23}.\frac{1}{9}=\frac{2}{23}.\frac{1}{1}=\frac{2}{23}\\ 8.\frac{4}{3}.\frac{9}{4}=8.\frac{1}{3}.\frac{9}{1}=\frac{8}{3}.9=\frac{72}{3}=4\\ mìnhchỉghicáchlàmkhôngghiđềbàinha\)
\(x+0,25=\dfrac{43}{4}-\dfrac{18}{5}\\ x+0,25=7,15\\ x=7,15-0,25\\x=6,9\)
Vậy x = 6,9
x = 69/10