Toán Lớp 8: quy đồng mẫu thức của các phân thức
1 /x+2; -3x/x-2; 3/x^2-4x+4
-1/2x+2; 3/2-2x;5/4x^2+4x+1
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Toán Lớp 8: quy đồng mẫu thức của các phân thức
1 /x+2; -3x/x-2; 3/x^2-4x+4
-1/2x+2; 3/2-2x;5/4x^2+4x+1
Comments ( 1 )
a)\dfrac{1}{{x + 2}};\dfrac{{ – 3x}}{{x – 2}};\dfrac{3}{{{x^2} – 4x + 4}}\\
MSC = {\left( {x – 2} \right)^2}.\left( {x + 2} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + 2}} = \dfrac{{{{\left( {x – 2} \right)}^2}}}{{{{\left( {x – 2} \right)}^2}\left( {x + 2} \right)}}\\
\dfrac{{ – 3x}}{{x – 2}} = \dfrac{{ – 3x\left( {x – 2} \right)\left( {x + 2} \right)}}{{{{\left( {x – 2} \right)}^2}\left( {x + 2} \right)}} = \dfrac{{ – 3{x^3} + 12x}}{{{{\left( {x – 2} \right)}^2}\left( {x + 2} \right)}}\\
\dfrac{3}{{{x^2} – 4x + 4}} = \dfrac{{3\left( {x + 2} \right)}}{{{{\left( {x – 2} \right)}^2}\left( {x + 2} \right)}} = \dfrac{{3x + 6}}{{{{\left( {x – 2} \right)}^2}\left( {x + 2} \right)}}
\end{array} \right.\\
b)\\
\dfrac{{ – 1}}{{2x + 2}};\dfrac{3}{{2 – 2x}};\dfrac{5}{{4{x^2} + 4x + 1}}\\
MSC = 2.\left( {x + 1} \right)\left( {1 – x} \right).{\left( {2x + 1} \right)^2}\\
+ ) – \dfrac{1}{{2x + 2}} = \dfrac{{ – \left( {1 – x} \right){{\left( {2x + 1} \right)}^2}}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{\left( {x – 1} \right)\left( {4{x^2} + 4x + 1} \right)}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{4{x^3} – 3x – 1}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
+ )\dfrac{3}{{2 – 2x}} = \dfrac{{3\left( {x + 1} \right){{\left( {2x + 1} \right)}^2}}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{\left( {3x + 3} \right)\left( {4{x^2} + 4x + 1} \right)}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{12{x^3} + 24{x^2} + 15x + 3}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
+ )\dfrac{5}{{4{x^2} + 4x + 1}} = \dfrac{{5.2.\left( {x + 1} \right)\left( {1 – x} \right)}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{10\left( {1 – {x^2}} \right)}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{10 – 10{x^2}}}{{2.\left( {x + 1} \right)\left( {1 – x} \right).{{\left( {2x + 1} \right)}^2}}}
\end{array}$