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a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\Rightarrow\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\Rightarrow\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+329=0\\\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}=0\left(vôlí\right)\end{matrix}\right.\)
\(\Rightarrow x=-329\)
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
⇔ \(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\)
\(\left(\dfrac{x+349}{5}-4\right)=0\)
⇔ \(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
⇔ \(\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
⇔ \(x+329=0\) Vì \(\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)\) ≠ 0
⇔ \(x=-329\)
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\\ \Rightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\\ \Rightarrow\dfrac{x-100}{30}+\dfrac{x-100}{43}+\dfrac{x-100}{95}+\dfrac{x-100}{8}=0\\ \Rightarrow x-100=0\\ \Rightarrow x=100\)
a) \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\right)=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\) \(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\\ \Rightarrow\dfrac{x}{2008}=1\\ \Rightarrow x=2008\)
b) \(\dfrac{7}{x}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}=\dfrac{29}{45}\)
\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}\right)=\dfrac{29}{45}\)
\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)
\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{7}{15}\\ \Rightarrow x=15\)
c) \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)
\(\Rightarrow2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=\dfrac{15}{93}.2\)
\(\Rightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\\ \Rightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\\ \Rightarrow\dfrac{2x}{3\left(2x+3\right)}=\dfrac{10}{31}\\ \Rightarrow\dfrac{10.3\left(2x+3\right)}{31}=2x\\ \Rightarrow\dfrac{30\left(2x+3\right)}{31}=2x\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{31}:2\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{62}\\ \Rightarrow x=\dfrac{15\left(2x+3\right)}{31}\\\Rightarrow\dfrac{15\left(2x+3\right)}{x}=31\\ \Rightarrow\dfrac{30x+45}{x}=31\\ \Rightarrow30+\dfrac{45}{x}=31\\ \Rightarrow \dfrac{45}{x}=1\\ \Rightarrow x=45\)
a/ \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-............-\dfrac{1}{120}=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+.......+\dfrac{1}{120}\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+.......+\dfrac{2}{240}\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}-\dfrac{3}{16}=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{x}{2008}=\dfrac{13}{16}\)
\(\Leftrightarrow x=1631,5\)
Vậy ..................
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
Câu 2:
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\)
\(\Leftrightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\)
=>x-100=0
hay x=100
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 4 = 0
(1 + \(\dfrac{55-x}{1963}\) ) + ( 1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0
\(\dfrac{1963+55-x}{1963}\) + \(\dfrac{1968+50-x}{1968}\)+\(\dfrac{1973+45-x}{1973}\)+\(\dfrac{1978+40-x}{1978}\)=0
\(\dfrac{2018-x}{1963}\)+\(\dfrac{2018-x}{1968}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1978}\)=0
(2018 - \(x\))\(\times\)( \(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0
2018 \(-x\) = 0
\(x\) = 2018
Giải:
\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903-x}{97}+4=0\)
\(\Leftrightarrow\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903-x}{97}+1=0\)
\(\Leftrightarrow\dfrac{1909-x+91}{91}+\dfrac{1907-x+93}{93}+\dfrac{1905-x+95}{95}+\dfrac{1903-x+97}{97}=0\)
\(\Leftrightarrow\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)
\(\Leftrightarrow\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)
Vì \(\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)\ne0\)
Nên \(2000-x=0\)
\(\Leftrightarrow x=2000\)
Vậy \(x=2000\)