$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

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

$\Rightarrow \left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)$

Vì 10<11<12<13<14 $\Rightarrow \frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}$

$\Rightarrow \frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

 Câu 1:x+1/10 + x+1/11 = x+1/12 + x+1/13 + x+1/14.

<-> (x+1)(1/10+1/11-1/12-1/13-1/14)=0

<-> x+1=0

<-> x=-1

Câu 2:

x+4/2000+x+3/2001=x+2/2002+x

⇔x+4/2000+1+x+3/2001=x+2/2002+1+x+1/2003

⇔x+2004/2000+x+2004/2001=x+2004/2002+x+2004/2003

⇔(x+2004)/(1/2000+1/2001−1/2002−1/2003)=0

⇔x+2004=0

⇔x=-2004

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) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)

\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)

\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)

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) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)           \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

\(\Leftrightarrow x=-2004\)

Sorry mink mới lớp 5 nên ko thể giúp bn lm bài toán này thành thật xin lỗi 

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong ngoặc thứ hai \(\ne\)0

Do đó \(x+1=0\)\(\Rightarrow x=0-1=-1\)

b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+4}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)nên biểu thức trong ngoặc thứ hai phải \(\ne\)0

Do đó \(x+2004=0\)\(\Rightarrow x=0-2004=-2004\)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)                                            (*)

Vì \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\) nên \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)(**)

Từ (*) và (**) ta co \(x+1=0\Rightarrow x=-1\)

b) Làm tương tự nha bạn \(x=-4\)

a) <=>(x+1)(1/10 + 1/11+1/12) =(x+1)(1/13 + 1/14) 
<=>(x+1)(1/10 + 1/11+1/12 -1/13 -1/14)=0 
<=> x+1=0(vì biểu thức 1/10 + 1/11 +1/12-1/13-1/14#0) 
<=>x= -1

b) (x+4)/2000 + (x+3)/2001 = (x+2)/2002 + (x+1)/2003 
<=> (x+4)/2000 + 1 + (x+3)/2001 +1 = (x+2)/2002 + 1 + (x+1)/2003 + 1 (thêm 2 vào mỗi vế ) 
<=> (x+4+2000)/2000 + (x+3+2001)/2001 = (x+2+2002)/2002 + (x+1+2003)/2003 
<=> (x+2004)/2000 + (x+2004)/2001 - (x+2004)/2002 - (x+2004)/2003 = 0 ( chuyển vế ) 
<=> (x+2004)(1/2000 + 1/2001 - 1/2002 - 1/2003) = 0 ( nhóm hạng tử x + 2004) 
vậy biể thức trên bằng 0 tại x+2004 = 0 hoặc 1/2000 + 1/2001 - 1/2002 - 1/2003 = 0 
mà ta dễ thấy 1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0 
nên biểu thức trên bằng 0 tại x+2004=0 
=> x = -2004 
vậy S = { -2004}

a/ \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Vì: \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

Vậy:.....

Tại sao lại =0??

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

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

Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

nên x + 1 = 0 => x = -1

Vậy x = -1

b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)

\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)

\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)

\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)

nên 2004 + x = 0 => x = -2004

Vậy x = -2004

=))

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