Toán Lớp 8: Quy đồng các phân thức
1/ x^2-x ; 3/x^2-1 ; 2/x^2+2x+1
Mọi người giúp e vs!!!!
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Toán Lớp 8: Quy đồng các phân thức
1/ x^2-x ; 3/x^2-1 ; 2/x^2+2x+1
Mọi người giúp e vs!!!!
Comments ( 1 )
\dfrac{{{x^2} + 2x + 1}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}\\
\dfrac{{3{x^2} + 3x}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}\\
\dfrac{{2{x^2} – 2x}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}
\end{array}\)
\dfrac{1}{{{x^2} – x}} = \dfrac{1}{{x\left( {x – 1} \right)}} = \dfrac{{{x^2} + 2x + 1}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}\\
\dfrac{3}{{{x^2} – 1}} = \dfrac{{3x\left( {x + 1} \right)}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}} = \dfrac{{3{x^2} + 3x}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}\\
\dfrac{2}{{{x^2} + 2x + 1}} = \dfrac{{2x\left( {x – 1} \right)}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}} = \dfrac{{2{x^2} – 2x}}{{x\left( {x – 1} \right){{\left( {x + 1} \right)}^2}}}
\end{array}\)