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để ý 1+1/x(x+2)=(x2+2x+1)/x(x+2)=(x+1)2/x(x+2)
+ 1+1/1.3=22/1.3 ;......
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(=>x^2+x+1-3x^2=2x\left(x-1\right)\)
\(=>-2x^2+x+1=2x^2-2x\)
\(=>-4x^2+3x+1=0\)
\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'
\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)
\(\left(\frac{3}{1.3}+\frac{3}{3.5}+.......+\frac{3}{97.99}\right).\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{97.99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{97}-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\left(\frac{3}{2}.\frac{98}{99}\right).\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\frac{49}{33}.\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\frac{49}{33}.2x+\frac{49}{33}=x+\frac{1}{33}\)
\(\Rightarrow\frac{98}{33}.x+\frac{49}{33}=x+\frac{1}{33}\)
\(\Rightarrow\frac{98}{33}.x-x=\frac{1}{33}-\frac{49}{33}\)
\(\Rightarrow\frac{65}{33}.x=\frac{-16}{11}\)
\(\Rightarrow x=\frac{-16}{11}:\frac{65}{33}\)
\(\Rightarrow x=\frac{-48}{65}\)
Vậy \(x=\frac{-48}{65}\)