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\(\dfrac{x+32}{11}+\dfrac{x+33}{12}=\dfrac{x+34}{13}+\dfrac{x+35}{14}\)
\(\Leftrightarrow\left(\dfrac{x+32}{11}-1\right)+\left(\dfrac{x+33}{12}-1\right)=\left(\dfrac{x+34}{13}-1\right)+\left(\dfrac{x+35}{14}-1\right)\)
\(\Leftrightarrow\dfrac{x+21}{11}+\dfrac{x+21}{12}=\dfrac{x+21}{13}+\dfrac{x+21}{14}\)
\(\Leftrightarrow\dfrac{x+21}{11}+\dfrac{x+21}{12}-\dfrac{x+21}{13}-\dfrac{x+21}{14}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Mà \(\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Leftrightarrow x+21=0\)
\(\Leftrightarrow x=-21\)
Vậy ..
Tìm x, biết:
\(\dfrac{x-9}{11}+\dfrac{x-10}{12}+\dfrac{x-11}{13}=\dfrac{x-12}{14}+\dfrac{x-28}{15}\)
\(\Leftrightarrow\left(\dfrac{x-9}{11}+1\right)+\left(\dfrac{x-10}{12}+1\right)+\left(\dfrac{x-11}{13}+1\right)=\left(\dfrac{x-12}{14}+1\right)+\left(\dfrac{x-28}{15}+2\right)\)
=>x+2=0
=>x=-2
Có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Leftrightarrow\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Dấu "=" xảy ra:
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}=0\end{matrix}\right.\)
Vì \(\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)\ne0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=0+2=2\)
Vậy \(x=2\).
Học tốt!
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{1}{11}+\dfrac{1}{12}\right)\left(x+2\right)+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{23\left(x+2\right)}{132}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{23}{132}+\dfrac{1}{13}\right)\left(x+2\right)=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\left(x+2\right)\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{29\left(x+2\right)}{210}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}-\dfrac{29\left(x+2\right)}{210}=0\)
\(\Rightarrow\left(\dfrac{431}{6.286}-\dfrac{29}{6.35}\right)\left(x+2\right)=0\)
\(\Rightarrow\dfrac{1}{6}\left(\dfrac{431}{286}-\dfrac{29}{35}\right)\left(x+2\right)=-2\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
=> x+1=0
<=> x=-1
b) \(\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
<=> \(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
đến đây tương tự a
a) Ta có:
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(Vì:\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
Vậy....
b)Sửa lại đề nha
Ta có:
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
Lý giải tương tự câu a và kết luận nha
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+1\right)\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
<=> \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}-\dfrac{x+1}{14}-\dfrac{x+1}{15}=0\)
<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Do: \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{14}>0\) nên x + 1 = 0
Vậy x = -1
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
x-2/11 + x-2/12 +x-2/13 = x-2/14 + x-2/15
=> x-2 /11 + x-2/12 +x-2/13 - x-2/14 - x-2/15 = 0
=> (x-2). ( 1/11 + 1/12 + 1/13 - 1/14-1/15) = 0
=> x-2 = 0 => x=2
1/ 11 + 1/12 +1/13 -1/14 - 1/15 = 0
Vì 1/11; 1/12; 1/13; 1/14; 1/15 > 1 nên 1/11+1/12+1/3-1/14-1/15= 0 (vô lí)
Vậy x=2
Nhớ like
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.