(m:1-m x 1):(m x 1999 + m +1)
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13 tháng 7 2019
a) Có vẻ đề o đúng lắm . Theo mình o phải là 11/11 mà 1/11
Ta có \(\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>...>\frac{1}{19}>\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
hay \(S>\frac{1}{2}\)
b)Ta có 1998 x 1999 + 3997=(2000-2) x 1999 +3997 = 2000 x 1999 - 2 x 1999 +3997 = 1999 x 2000 -3998 +3997 =1999 x 2000 -1
< 1999 x 2000 +2
=> 1999 x 2000 +2 / 1998 x 1999 +3997 > 1 hay M>1
NT
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23 tháng 3 2017
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1999}{2001}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1999}{2001}\)
\(\Leftrightarrow2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1999}{2001}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1999}{4002}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1999}{4002}\)\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2001}\)
\(\Leftrightarrow x+1=2001\Leftrightarrow x=2000\)
NT
1
21 tháng 4 2017
Câu 1:
Tại \(x=5\) thì ta có pt:
\(pt\Leftrightarrow10+4m^2=19\)
\(\Leftrightarrow4m^2=9\Leftrightarrow m^2=\dfrac{9}{4}\)
\(\Leftrightarrow m=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\)
Vậy với \(m=\pm\dfrac{3}{2}\) thì pt có nghiệm là \(x=5\)
Câu 2:
\(\dfrac{x+5}{1999}+\dfrac{x+7}{1997}=\dfrac{x+9}{1995}+\dfrac{x+11}{1993}\)
\(\Leftrightarrow\dfrac{x+5}{1999}+1+\dfrac{x+7}{1997}+1=\dfrac{x+9}{1995}+1+\dfrac{x+11}{1993}+1\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}=\dfrac{x+2004}{1995}+\dfrac{x+2004}{1993}\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}-\dfrac{x+2004}{1995}-\dfrac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\right)=0\)
\(\Rightarrow x+2004=0\). Do \(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\ne0\)
\(\Rightarrow x=-2014\)
7 tháng 6 2019
a, (sửa đề )
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2000}\)
=\(1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{1999}{2000}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x+\left(x+1\right)}=1-\frac{1999}{2000}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{x+1}=\frac{1}{1}-\frac{1}{2000}=\frac{1999}{2000}\)
=> \(x+1=1:\frac{1999}{2000}=\frac{2000}{1999}\)
=>\(x=\frac{2000}{1999}-1=\frac{1}{1999}\)
Vậy x ∈{ \(\frac{1}{1999}\)}
b, \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>2.(\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2}{9}\)
=>\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x+\left(x+1\right)}=\frac{2}{9}:2=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=>\(x+1=18\)
=>\(x=18-1=17\)
=>x∈{17}
24 tháng 5 2022
\(\Leftrightarrow\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{1999}{1993}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3992}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-3}{1993}\)
=>x+1=-1993/3
hay x=-1996/3
23 tháng 7 2017
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+1999\right)\left(x+2000\right)}=\dfrac{1}{x+200}=\dfrac{1}{5}\)\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+1999}-\dfrac{1}{x-2000}=\dfrac{1}{x+200}=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2000}=\dfrac{1}{x+200}=\dfrac{1}{5}\)
Đề này sai nhé,hình như thừa dữ kiện đề r
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