Tìm x
\(\dfrac{x+32}{11}+\dfrac{x+23}{12}=\dfrac{x+38}{13}+\dfrac{x+27}{14}\)
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L
1
28 tháng 6 2018
\(\dfrac{x+32}{11}+\dfrac{x+33}{12}=\dfrac{x+34}{13}+\dfrac{x+35}{14}\)
\(\Leftrightarrow\left(\dfrac{x+32}{11}-1\right)+\left(\dfrac{x+33}{12}-1\right)=\left(\dfrac{x+34}{13}-1\right)+\left(\dfrac{x+35}{14}-1\right)\)
\(\Leftrightarrow\dfrac{x+21}{11}+\dfrac{x+21}{12}=\dfrac{x+21}{13}+\dfrac{x+21}{14}\)
\(\Leftrightarrow\dfrac{x+21}{11}+\dfrac{x+21}{12}-\dfrac{x+21}{13}-\dfrac{x+21}{14}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Mà \(\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Leftrightarrow x+21=0\)
\(\Leftrightarrow x=-21\)
Vậy ..
O
9 tháng 5 2022
B1 :
a) \(\dfrac{2}{15}+3=\dfrac{2}{15}+\dfrac{15}{5}=\dfrac{17}{5}\)
b) \(2-\dfrac{7}{4}=\dfrac{8}{4}-\dfrac{7}{4}=\dfrac{1}{4}\)
c) \(\dfrac{17}{6}-\dfrac{11}{12}\times6=\dfrac{17}{6}-\dfrac{11}{2}=-\dfrac{8}{3}\)
d) \(\dfrac{14}{17}:7-\dfrac{15}{34}=\dfrac{2}{17}-\dfrac{5}{34}=\dfrac{4}{34}-\dfrac{5}{34}=-\dfrac{1}{34}\)
B3 :
\(x:23=146\)
\(x=146\times23\)
\(x=3358\)
b) \(x\times38=4066\)
\(x=4066:38\)
\(x=107\)
c) \(787+x\times67=2658\)
\(x\times67=2658-787\)
\(x\times67=1871\)
\(x=1871:67\)
\(x=\dfrac{1871}{67}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
24 tháng 4 2022
`9/14 : 3/7 + 5/6 = 3/98 + 5/6 = 127/147`
__________________
`4/7 xx (4 - 7/3) = 4/7 xx 5/3 = 20/21`
AD
4
5 tháng 7 2023
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)\(\left(x+1\right)\times\dfrac{1}{10}+\left(x+1\right)\times\dfrac{1}{11}+\left(x+1\right)\times\dfrac{1}{12}-\left(x+1\right)\times\dfrac{1}{13}-\left(x+1\right)\times\dfrac{1}{14}=0\)
\(\left(x+1\right)\times\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\)
=> \(x+1=0\)
\(x=0-1\)
\(x=-1\)
H
5 tháng 7 2023
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\\ \Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\\ \Rightarrow x+1=0\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\\ \Rightarrow x=-1\)
NL
1
AH
Akai Haruma
Giáo viên
16 tháng 9 2023
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 $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
NJ
Tìm x, biết:
\(\dfrac{x-9}{11}+\dfrac{x-10}{12}+\dfrac{x-11}{13}=\dfrac{x-12}{14}+\dfrac{x-28}{15}\)
1
22 tháng 11 2022
\(\Leftrightarrow\left(\dfrac{x-9}{11}+1\right)+\left(\dfrac{x-10}{12}+1\right)+\left(\dfrac{x-11}{13}+1\right)=\left(\dfrac{x-12}{14}+1\right)+\left(\dfrac{x-28}{15}+2\right)\)
=>x+2=0
=>x=-2
Lời giải:
\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
\(\Leftrightarrow \frac{x+32}{11}-3+\frac{x+23}{12}-2=\frac{x+38}{13}-3+\frac{x+27}{14}-2\)
\(\Leftrightarrow \frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)
\(\Leftrightarrow (x-1)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Dễ thấy: \(\frac{1}{11}+\frac{1}{12}> \frac{1}{13}+\frac{1}{14}\Rightarrow \frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\neq 0\)
Do đó: \(x-1=0\Leftrightarrow x=1\) là nghiệm duy nhất.