Toán Lớp 8: a. 3x/5x+5y-x/10x-10y
b. 1/3x-2-4/3x+2-3x-6/4-9x^2
c. 6x-4/x^2-4.x^2-3x+2/3x-2
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Toán Lớp 8: a. 3x/5x+5y-x/10x-10y
b. 1/3x-2-4/3x+2-3x-6/4-9x^2
c. 6x-4/x^2-4.x^2-3x+2/3x-2
Comments ( 1 )
a)\dfrac{{3x}}{{5x + 5y}} – \dfrac{x}{{10x – 10y}}\\
= \dfrac{{3x}}{{5\left( {x + y} \right)}} – \dfrac{x}{{10\left( {x – y} \right)}}\\
= \dfrac{{3.2\left( {x – y} \right) – x\left( {x + y} \right)}}{{10\left( {x + y} \right)\left( {x – y} \right)}}\\
= \dfrac{{6x – 6y – {x^2} – xy}}{{10\left( {{x^2} – {y^2}} \right)}}\\
b)\dfrac{1}{{3x – 2}} – \dfrac{4}{{3x + 2}} – \dfrac{{3x – 6}}{{4 – 9{x^2}}}\\
= \dfrac{{3x + 2 – 4\left( {3x – 2} \right) + 3x – 6}}{{\left( {3x + 2} \right)\left( {3x – 2} \right)}}\\
= \dfrac{{3x + 2 – 12x + 8 + 3x – 6}}{{\left( {3x + 2} \right)\left( {3x – 2} \right)}}\\
= \dfrac{{4 – 6x}}{{\left( {3x + 2} \right)\left( {3x – 2} \right)}}\\
= \dfrac{{ – 2\left( {3x – 2} \right)}}{{\left( {3x + 2} \right)\left( {3x – 2} \right)}}\\
= \dfrac{{ – 2}}{{3x + 2}}\\
c)\dfrac{{6x – 4}}{{{x^2} – 4}}.\dfrac{{{x^2} – 3x + 2}}{{3x – 2}}\\
= \dfrac{{2\left( {3x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 1} \right)\left( {x – 2} \right)}}{{3x – 2}}\\
= \dfrac{{2\left( {x – 1} \right)}}{{x + 2}}
\end{array}$