\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)

\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)

Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)

\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)

\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)

\(\Leftrightarrow\left(x+60\right)\text{=}0\)

\(x\text{=}-60\)

\(Vậy...\)

`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`

`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`

`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`

`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`

`<=> x * 88/315 = -352/45`

`<=> x = -28`

Vậy `S={-28}`

mình lười nên nói cách làm nhé

B1: chuyển \(\dfrac{6}{x^2-9}\)sang vế trái và thêm dấu trừ ở trc \(\dfrac{6}{x^2-9}\)và vế phải =0

B2: để ý thấy \(x^2-9\)=(x-3).(x+3) tức là hằng đẳng thức số 3 ý

B3: quy đồng mẫu , mẫu số chung là (x-3).(x+3).(2x+7)

B4: chia cả hai vế cho (x-3).(x+3).(2x+7)

lưu ý : bước này là dấu⇒ chứ ko phải dấu ⇔ nhé

B5: giải pt như bình thg thui

ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{7}{2}\right\}\)

Ta có: \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\)

\(\Leftrightarrow\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

Suy ra: \(13x+39+x^2-9=12x+42\)

\(\Leftrightarrow x^2+13x+30-12x-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-4}

\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)=72\left(x+3\right)+84\left(x+4\right)\)

\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)-72\left(x+3\right)-84\left(x+4\right)=0\)

\(\Leftrightarrow-37x-370=0\Leftrightarrow x=-10\)

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

Vậy $x = -10$

`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`

`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`

`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`

`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`

<=> x-2019=0`

`<=> x=2019`

\(\dfrac{x-2003}{16}+\dfrac{x-1997}{11}+\dfrac{x-1992}{9}+\dfrac{x-1991}{7}=10\)

\(\Leftrightarrow\left(\dfrac{x-2003}{16}-1\right)+\left(\dfrac{x-1997}{11}-2\right)+\left(\dfrac{x-1992}{9}-3\right)+\left(\dfrac{x-1991}{7}-4\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-2003}{16}-\dfrac{16}{16}\right)+\left(\dfrac{x-1997}{11}-\dfrac{22}{11}\right)+\left(\dfrac{x-1992}{9}-\dfrac{27}{9}\right)+\left(\dfrac{x-1991}{7}-\dfrac{28}{7}\right)=0\)

\(\Leftrightarrow\dfrac{x-2003-16}{16}+\dfrac{x-1997-22}{11}+\dfrac{x-1992-27}{9}+\dfrac{x-1991-28}{7}=0\)

\(\Leftrightarrow\dfrac{x-2019}{16}+\dfrac{x-2019}{11}+\dfrac{x-2019}{9}+\dfrac{x-2019}{7}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow x-2019=0\) Vì \(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\ne0\)

\(\Leftrightarrow x=2019\)

Vậy phương trình có 1 nghiệm \(x=2019\)

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)

Ta có :

\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)

Tương tự :

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

\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)

....

\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)

Từ những ý trên , pt trở thành :

\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)

\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)

\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)

\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)

\(\Leftrightarrow121x-900480=0\)

\(\Leftrightarrow x=\dfrac{900480}{121}\)