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(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)
Vậy x>9 thì (x-5)(x-9)>0
ta có:
\(\frac{x+2}{2013}+\frac{x+5}{2010}>\frac{x+8}{2007}+\frac{x+11}{2004}\)
\(\Leftrightarrow\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+5}{2010}+1\right)>\left(\frac{x+8}{2007}+1\right)+\left(\frac{x+11}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}>\frac{x+2015}{2007}+\frac{x+2015}{2004}\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}-\frac{x+2015}{2007}-\frac{x+2015}{2004}>0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}\right)>0\)
\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)
b, \(\frac{3x-2}{5}\ge\frac{x+1,6}{2}\)
=> \(6x-4\ge5x+8\)
=> \(x-12\ge0\)
=> \(x\ge12\)
bpt 2: \(\frac{6-2x+5}{6}>\frac{3-x}{4}\)
=> \(\frac{11-2x}{6}>\frac{3-x}{4}\)
=> \(44-8x>18-6x\)
=> \(x< 13\)
Vậy để t/m cả 2 bpt thì : \(12\le x< 13\)
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
a) \(\frac{5x-3}{5}+\frac{2x+1}{4}\le\frac{2-3x}{2}-5\)
\(\Leftrightarrow\frac{4\cdot\left(5x-3\right)}{20}+\frac{5\left(2x+1\right)}{20}\le\frac{10\left(2-3x\right)}{20}-\frac{20\cdot5}{20}\)
\(\Leftrightarrow20x-12+10x+5\le20-30x-100\)
\(\Leftrightarrow20x+10x+30x\le20-100+12-5\)
\(\Leftrightarrow60x\le-73\)
\(\Leftrightarrow x\le\frac{-73}{60}\)
a, x+2/5 >=0 <=> x+2 >=0 <=> x>=-2
b. x+2/x-3 <0 <=> 1+5/x-3 <0 <=> 5/x-3 <-1 <=> x-3> -5 <=> x>-2
c. x-1/x-3 >1 <=> 1+ 2/x-3 >1 <=> 2/x-3 >0 <=> x-3 >0 <=> x>3
a, \(\Rightarrow\)\(1+\frac{x+3}{2011}\)\(+1+\frac{x+1}{2013}\)\(\ge1+\frac{x+10}{2004}+1+\frac{x+13}{2001}\)
\(\Rightarrow\)\(\frac{2011+x+3}{2011}+\frac{2013+x+1}{2013}\ge\frac{2004+x+10}{2004}+\frac{2001+x+13}{2001}\)
\(\Rightarrow\)\(\frac{2014+x}{2011}+\frac{2014+x}{2013}\ge\frac{2014+x}{2004}+\frac{2014+x}{2001}\)
\(\Rightarrow\)\(\frac{2014+x}{2011}+\frac{2014+x}{2013}-\frac{2014+x}{2004}+\frac{2014+x}{2001}\ge0\)
\(\Rightarrow\)\(\left(2014+x\right)\left(\frac{1}{2011}+\frac{1}{2013}-\frac{1}{2004}-\frac{1}{2001}\right)\)\(\ge0\)
\(do\)\(\frac{1}{2011}+\frac{1}{2013}-\frac{1}{2004}-\frac{1}{2001}< 0\)
\(\Rightarrow\)\(2014+x\le0\)
\(\Rightarrow\)\(x\le-2014\)