$\frac{x+20}{2000}+\frac{x+45}{2005}+\frac{x-5}{2025}=\frac{x+45}{1975}+\frac{x=90}{2140}\frac{x-20}{2040}$
Những câu hỏi liên quan
Biết: \(\frac{x+4}{45^2-5^2}+\frac{x-46}{45^2+5^2}-\frac{x-96}{50^2-20^2}=\frac{3x+6012}{50^2+20^2}+\frac{x}{2004}\)
Tính: \(S=\frac{|x|^2-1}{x+1}\)
Biết: \(\frac{x+4}{45^2-5^2}+\frac{x-46}{45^2+5^2}-\frac{x-96}{50^2-20^2}=\frac{3x+6012}{50^2+20^2}+\frac{x}{2004}\)
Tính: \(S=\frac{|x|^2-1}{x+1}\)
Bài 9: Tìm x biết:
a, \(x+\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+....+\frac{4}{41\times45}=\frac{-37}{45}\)
b, \(x-\frac{20}{11\times13}-\frac{20}{13\times15}-\frac{20}{15\times17}-....-\frac{20}{53\times55}=\frac{3}{11}\)
c, \(\frac{1}{21}+\frac{1}{21}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
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\(\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{4034}{5}\)
\(x\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)=\frac{4034}{5}\)
\(x.\frac{2}{5}=\frac{4034}{5}\)
\(x=\frac{4034}{5}:\frac{2}{5}\)
\(x=\frac{4034}{5}.\frac{5}{2}\)
x = 2017
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Tìm số tự nhiên x, biết:
a, \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-....-\frac{20}{53-55}=\frac{3}{11}\)
b, \(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
c,\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
1. Tìm x\
a) Tìm x, biết :\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.47}=\frac{-37}{45}\)
b) Tìm x\(\in\)n, biết:\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
Giải bất phương trình sau:
\(\frac{x+20}{31}-\frac{x+27}{33}\ge\frac{x+17}{43}-\frac{x+15}{45}\)
Tìm x biết :
a/\(\frac{x+1945}{45}+\frac{x+1954}{54}\)=\(\frac{x+1975}{75}+\frac{x+1969}{69}\)
b/\(\frac{x-99}{5}+\frac{x-97}{7}\)=\(\frac{x-95}{9}+\frac{x-93}{11}\)
c/\(\frac{x+2008}{10}+\frac{x+2010}{9}\)=\(\frac{x+2012}{8}+\frac{x+2014}{7}\)
b) \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)
\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)=\left(\frac{x-95}{9}-1\right)\)\(+\left(\frac{x-93}{11}-1\right)\)
\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)
\(\Leftrightarrow\left(x-104\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)
Mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\ne0\)
\(\Rightarrow x-104=0\)
\(\Leftrightarrow x=104\)
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a) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)\)\(+\left(\frac{x+1969}{69}-1\right)\)
\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)
\(\Leftrightarrow\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
Mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\)
\(\Rightarrow x+1900=0\)
\(\Leftrightarrow x=-1900\)
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c) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)
\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)\)\(=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)
\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)
\(\Leftrightarrow\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\)
\(\Rightarrow x+2028=0\)
\(\Leftrightarrow x=-2028\)
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a. \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\)
b.\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
c.\(\frac{7x-1}{2}=5+\frac{9-5x}{6}\)
d.\(\frac{3x-9}{x+1}-2=\frac{4x}{x+1}\)
c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)
\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)
\(\Rightarrow42x-6=60+18-10x\)
\(\Leftrightarrow52x-84=0\)
\(\Leftrightarrow x=\dfrac{21}{13}\)
Vậy....
d) tương tự
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a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
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b) \(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}-1\)
\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}-1+1+1\)
\(\Leftrightarrow\dfrac{x+16+49}{49}+\dfrac{x+18+47}{47}=\dfrac{x+20+45}{45}\)
\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)
\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)
Vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
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