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$\dfrac{x+32}{11}+\dfrac{x+23}{12}=\dfrac{x+38}{13}+\dfrac{x+27}{14}$

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Akai Haruma · Accepted Answer

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.Câu trả lời: 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.