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75.174 - 75.75 + 75
= 75. (174 - 75) + 75
= 75. 99 + 75
= 7425 + 75
= 7500
\(71+52,5\times4=\frac{x+140}{x}+210\)
\(71+210=\frac{x+140}{x}+210\)
\(=>\frac{x+140}{x}=71\)
\(71=\frac{142}{2}\)\(\Rightarrow x=142-140=2\)
Chuyển dấu phẩy của số đó sang phải 1 hàng thì số đó gấp lên 10 lần
Số cần tìm là:
178.65:(10-1)x1=19.85
Đáp số:19.85
<=> 71=\(\dfrac{x+140}{x}\)
=>71x=x+140
<=>71x-x=140
<=>70x=140
<=>x=2
Vậy x=2
\(71+65\cdot4=\frac{x+140}{x}+260\)
\(\Rightarrow71+260=\frac{x}{x}+\frac{140}{x}+260\)
\(\Rightarrow71=1+\frac{140}{x}\)
\(\Rightarrow\frac{140}{x}=70\)
\(\Rightarrow x=2\)
71+65*4=x+140 +260
x
=>71+260= x + 140 +260
x x
=>71=1+ 140
x
=> 71-1 = 140
x
=> 70=140:x
=>x=140:70=2
=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 = \(\dfrac{502}{3}\)
Theo đầu bài ta có:
\(\frac{x-7}{7}+\frac{x-7}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{1}{7}+\left(x-7\right)\cdot\frac{1}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\left(\frac{1}{7}+\frac{1}{8}\right)=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{15}{56}=\frac{15}{56}\)
\(\Rightarrow x-7=1\)
\(\Rightarrow x=8\)
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
Lời giải:
$56+\frac{x+60,2}{x}=71$
$56+\frac{x}{x}+\frac{60,2}{x}=71$
$56+1+\frac{60,2}{x}=71$
$57+\frac{60,2}{x}=71$
$\frac{60,2}{x}=71-57=14$
$x=60,2:14=4,3$
Lời giải:
56+\frac{x+60,2}{x}=7156+xx+60,2=71
56+\frac{x}{x}+\frac{60,2}{x}=7156+xx+x60,2=71
56+1+\frac{60,2}{x}=7156+1+x60,2=71
57+\frac{60,2}{x}=7157+x60,2=71
\frac{60,2}{x}=71-57=14x60,2=71−57=14
x=60,2:14=4,3x=60,2:14=4,3