Lời giải:
$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12})=(x+1)(\frac{1}{13}+\frac{1}{14})$
$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14})=0$
Hiển nhiên $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
\(\dfrac{x+1}{10}\)+\(\dfrac{x+1}{11}\)+\(\dfrac{x+1}{12}\)=\(\dfrac{x+1}{13}\)+\(\dfrac{x+1}{14}\)
MỌI NGƯỜI GIÚP EM VỚI
Bài 1: tìm x
a)\(\left|3x-5\right|=4\)
b)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
c)\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
Bài 2: Tính
a)\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
b)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c)\(\dfrac{\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right).\dfrac{5}{19}}{\left(\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{-3}{35}\right).\dfrac{-4}{3}}\)
Tìm x, biết:
a) \(\dfrac{-1}{10}\) + \(\dfrac{2}{5}\)x + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)
b) \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) : x= \(-\dfrac{1}{5}\)
c) \(-\dfrac{2}{3}\) : x + \(\dfrac{5}{8}\) = \(-\dfrac{7}{12}\)
1, Tính hợp lí
\(A=\dfrac{0.5+\dfrac{7}{12}-\dfrac{5}{6}}{1-\dfrac{2}{3}+0,75}\)
\(B=-66.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124.\left(-37\right)+126.\left(-62\right)\)
\(N=\left(60\dfrac{7}{13}+50\dfrac{8}{13}-11.\dfrac{2}{13}\right)x\) (Với \(x=-2017\dfrac{7}{10}\))
\(M=\left(1-\dfrac{2}{2.3}\right).\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right).....\left(1-\dfrac{2}{99.100}\right)\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
Tìm x:
\(\dfrac{x-2}{11}+\dfrac{x-2}{12}+\dfrac{x-2}{13}=\dfrac{x-2}{14}+\dfrac{x-2}{15}\)
tìm x \(\in\) Q biết rằng
\(\dfrac{11}{12}\) - ( \(\dfrac{2}{5}\) + x ) = \(\dfrac{2}{3}\)
2x \(\times\) ( x - \(\dfrac{1}{7}\) ) = 0
\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
Tìm x :
a, \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) +x) = \(\dfrac{2}{3}\) b, x+\(\dfrac{2}{3}\) =\(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\) c, x-[\(\dfrac{17}{2}\) -(\(\dfrac{-3}{7}\) + \(\dfrac{5}{3}\) )]=\(\dfrac{-1}{3}\)
d,\(\dfrac{9}{2}\) - [\(\dfrac{2}{3}\) - (x+\(\dfrac{7}{4}\) )] =\(\dfrac{-5}{4}\)
Mọi người giúp mình nha
Tìm x:
a) \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) + \(\dfrac{3}{4}\)x) \(\dfrac{2}{3}\)
b) \(\dfrac{-2}{5}\) + \(\dfrac{5}{3}\) . (\(\dfrac{3}{2}\) - \(\dfrac{4}{15}\)x) = \(\dfrac{-7}{6}\)
c) \(\dfrac{1}{2}\) + \(\dfrac{3}{4}\)x = \(\dfrac{1}{4}\)
