Câu hỏi của Nguyễn Bá Hải

Question

Tìm x:

a) $\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

b) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac...

Trần Thị Hương · Accepted Answer

a. $\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

Vì $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0$

$\Rightarrow x+1=0$

$x=-1$

b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\ \left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\ \dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\ x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)$

vì $\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\
\Rightarrow x+2004=0\
x=-2004$Câu trả lời: a. $\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

Vì $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0$

$\Rightarrow x+1=0$

$x=-1$

b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\ \left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\ \dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\ x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)$

vì $\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\
\Rightarrow x+2004=0\
x=-2004$

Đức Hiếu · Answer

a, $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$ $\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$ $\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$ Vì $10< 11< 12< 13< 14$ nên $\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}$ $\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0$ $\Rightarrow x+1=0\Rightarrow x=-1$ Vậy................. b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$ $\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)$ $\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$ $\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$ $\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$ Vì $2000< 2001< 2002< 2003$ nên $\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}$ $\Rightarrow\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}>0$ $\Rightarrow x+2004=0\Rightarrow x=-2004$ Vậy................. Chúc bạn học tốt!!!Câu trả lời: a, $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$ $\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$ $\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$ Vì $10< 11< 12< 13< 14$ nên $\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}$ $\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0$ $\Rightarrow x+1=0\Rightarrow x=-1$ Vậy................. b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$ $\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)$ $\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$ $\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$ $\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$ Vì $2000< 2001< 2002< 2003$ nên $\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}$ $\Rightarrow\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}>0$ $\Rightarrow x+2004=0\Rightarrow x=-2004$ Vậy................. Chúc bạn học tốt!!!

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