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\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)
\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)
\(\Leftrightarrow123-x=0\Leftrightarrow x=123\)
Vậy x = 123
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
\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Rightarrow\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3+\dfrac{x-166}{23}-4=0\)
\(\Rightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)
\(\Rightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)
Mà \(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0\)
\(\Rightarrow x-258=0\Rightarrow x=258\)
Vậy x = 258
x−24117+x−22019+x−19521+x−16623=10x−24117+x−22019+x−19521+x−16623=10
⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0
⇒x−25817+x−25819+x−25821+x−25823=0⇒x−25817+x−25819+x−25821+x−25823=0
⇒(x−258)(117+119+121+123)=0⇒(x−258)(117+119+121+123)=0
Mà 117+119+121+123≠0117+119+121+123≠0
⇒x−258=0⇒x=258
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
a) Ta có:
\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)
\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)
\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)
Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.
Vậy A > C > B.
b) Ta có:
\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)
\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)
Vậy B : A = -4
