\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)'

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)

\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\left(vì:\frac{1}{2009}+\frac{1}{2007}< \frac{1}{2005}+\frac{1}{2003}\right)\)

ta có (x+1/2009 +1) + ( x+3/2007 + 1)- (x+5/2005 +1) - (x+7/1993 + 1) = 0

=>(x +100/ 2009) + (x+100/2007) - (x+100/2005)-(x+100/1993)

=> (x +100) * (1/2009 + 1/2007+ 1/2005 + 1/1993) = 0

=> x = -100

Bạn cứ tinh ý để ý đến phần tử và mẫu cộng lại bằng 100. Khi bạn bỏ phần x + 100 ra thì còn lại như trên. Sau đó lược bỏ còn lại x = -100

Mạn phép mk không chép đề , mk làm luôn nhé

\(\dfrac{x+1}{2009}+1+\dfrac{x+3}{2007}+1=\dfrac{x+5}{2005}+1+\dfrac{x+7}{1993}+1\)

⇔ \(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2005}-\dfrac{x+2010}{1993}=0\)

⇔( x + 2010 )\(\left(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}\right)=0\)

Ta thấy : \(\dfrac{1}{2009}< \dfrac{1}{2007}< \dfrac{1}{2005}< \dfrac{1}{1993}\)

⇒ \(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}< 0\)

⇒ x + 2010 = 0

⇒ x = -2010

KL....

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:

\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)

\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý

b.

Đặt \(x+3=t\) 

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)

\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)

\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)

Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.

Ta có: \(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)

\(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+3}{2007}+1=\frac{x+5}{2005}+1+\frac{x+7}{2003}+1\)

\(\Leftrightarrow\frac{x+1+2009}{2009}+\frac{x+3+2007}{2007}=\frac{x+5+2005}{2005}+\frac{x+7+2003}{2003}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}-\frac{x+2010}{2005}-\frac{x+2010}{2003}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)

=> x + 2010 = 0

=> x             = -2010

Vậy x = -2010

\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)

\(\Leftrightarrow\left(\frac{x+2010}{2009}\right)+\left(\frac{x+2010}{2007}\right)=\left(\frac{x+2010}{2005}\right)+\left(\frac{x+2010}{2003}\right)\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2010=0\) ( Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\))

\(\Leftrightarrow x=-2010\)

Vậy tập nghiệm của phương trình là S = { -2010 } .

\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)

\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)

\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)

                                    |________________A________________|

Do A > 0

nên x - 2014 = 0

<=> x = 2014

Ta có : 

\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)

\(\Leftrightarrow\)\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)

\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)

\(\Leftrightarrow\)\(\left(x-2014\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)

Nên \(x-2014=0\)

\(\Rightarrow\)\(x=2014\)

Vậy \(x=2014\)

Chúc bạn học tốt ~ 

\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)

Trừ cả 2 vế cho 2 ta được :

\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)

\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)

\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)

\(\Leftrightarrow\left(x-2014\right)\times\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)

Mà : \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)

<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)

<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0

<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0

<=> x+2011=0

<=> x=-2011

Vậy pt có nghiệm là x=-2011

\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)

\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}-4=0\)

\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-7}{2007}-1\right)+\left(\dfrac{x-9}{2005}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2007}+\dfrac{x-2014}{2005}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\right)=0\)

\(\Leftrightarrow x-2014=0\) ( do \(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\ne0\))

\(\Leftrightarrow x=2014\)

Vậy phương trình có nghiệm S=\(\left\{2014\right\}\)