Toán Lớp 8: Quy đồng mẫu của các phân thức sau
A) 5/3x-12 và 3/2x² – 8x
B) 2x/5x² – 5x và x+3/x² – 1
C) 3x+1/x²-4 và 3/5x+10
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Toán Lớp 8: Quy đồng mẫu của các phân thức sau
A) 5/3x-12 và 3/2x² – 8x
B) 2x/5x² – 5x và x+3/x² – 1
C) 3x+1/x²-4 và 3/5x+10
Comments ( 1 )
a)\dfrac{{10x}}{{6x\left( {x – 4} \right)}}\\
\dfrac{9}{{6x\left( {x – 4} \right)}}\\
b)\dfrac{{2x + 2}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}}\\
\dfrac{{5x + 15}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}}\\
c)\dfrac{{15x + 5}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}\\
\dfrac{{3x – 6}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}
\end{array}\)
a)\dfrac{5}{{3x – 12}} = \dfrac{5}{{3\left( {x – 4} \right)}} = \dfrac{{5.2x}}{{6x\left( {x – 4} \right)}}\\
= \dfrac{{10x}}{{6x\left( {x – 4} \right)}}\\
\dfrac{3}{{2{x^2} – 8x}} = \dfrac{3}{{2x\left( {x – 4} \right)}} = \dfrac{{3.3}}{{6x\left( {x – 4} \right)}}\\
= \dfrac{9}{{6x\left( {x – 4} \right)}}\\
b)\dfrac{{2x}}{{5{x^2} – 5x}} = \dfrac{{2x}}{{5x\left( {x – 1} \right)}} = \dfrac{2}{{5\left( {x – 1} \right)}}\\
= \dfrac{{2\left( {x + 1} \right)}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{{2x + 2}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}}\\
\dfrac{{x + 3}}{{{x^2} – 1}} = \dfrac{{5\left( {x + 3} \right)}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{{5x + 15}}{{5\left( {x – 1} \right)\left( {x + 1} \right)}}\\
c)\dfrac{{3x + 1}}{{{x^2} – 4}} = \dfrac{{3x + 1}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{5\left( {3x + 1} \right)}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{15x + 5}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}\\
\dfrac{3}{{5x + 10}} = \dfrac{3}{{5\left( {x + 2} \right)}} = \dfrac{{3\left( {x – 2} \right)}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{3x – 6}}{{5\left( {x – 2} \right)\left( {x + 2} \right)}}
\end{array}\)