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a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)
=>x+2006=0
=>x=-2006
b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)
=>x-105=0
=>x=105
a) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
b) \(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}=\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+8}{1996}+1=\dfrac{x+12}{1992}+1+\dfrac{x+16}{1988}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{1996}-\dfrac{x+2004}{1992}-\dfrac{x+2004}{1988}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì \(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\ne0\))
\(\Leftrightarrow x=-2004\)
ĐKXĐ \(x\ne8;x\ne11;x\ne9;x\ne10\)
\(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)
\(\Leftrightarrow\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)
\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}\)
\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)
1) x=0
2) \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)
\(\Leftrightarrow\dfrac{x-11+x-8}{\left(x-8\right)\left(x-11\right)}-\dfrac{x-10+x-9}{\left(x-9\right)\left(x-10\right)}=0\)
\(\Leftrightarrow\dfrac{2x-19}{\left(x-8\right)\left(x-11\right)}=\dfrac{2x-19}{\left(x-9\right)\left(x-10\right)}\)
\(\Leftrightarrow\dfrac{2x-19}{x^2-19x+88}=\dfrac{2x-19}{x^2-19x+90}\)
do \(x^2-19x+88\ne x^2-19x+90\)
\(\Rightarrow2x-19=0\)
=> x=\(\dfrac{19}{2}\)
Vậy x=\(0\); x=\(\dfrac{19}{2}\)
Tik 
\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)
\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)
Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)
\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)
\(\Leftrightarrow\left(x+60\right)\text{=}0\)
\(x\text{=}-60\)
\(Vậy...\)
1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)
=>x+86=0
=>x=-86
2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)
=>x+2014=0
=>x=-2014
3: \(\Leftrightarrow3\left(x+4\right)-2\left(x-3\right)=4x\)
=>4x=3x+12-2x+6
=>4x=x+18
=>3x=18
=>x=6
4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)
=>4x-14+5x+55=-40
=>9x+41=-40
=>x=-9
`(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}`
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 \(x>0\)
điều kiện xác định \(x\ne0\)
ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)
\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)
tới đây bn bấm máy tính nha
`(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`