Toán Lớp 8: 1. Tìm x biết 10x^3-x^2-3x=0
2. Rút gọn phân thức: A= x^2-25/x^2+10x+25
B= x^3-27/x^3+3x^2+9x
C= 3x^2-12/6x^2+24x+24
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Toán Lớp 8: 1. Tìm x biết 10x^3-x^2-3x=0
2. Rút gọn phân thức: A= x^2-25/x^2+10x+25
B= x^3-27/x^3+3x^2+9x
C= 3x^2-12/6x^2+24x+24
Comments ( 1 )
1)10{x^3} – {x^2} – 3x = 0\\
\Leftrightarrow x\left( {10{x^2} – x – 3} \right) = 0\\
\Leftrightarrow x\left( {10{x^2} + 5x – 6x – 3} \right) = 0\\
\Leftrightarrow x.\left( {2x + 1} \right)\left( {5x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x + 1 = 0\\
5x – 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = – \dfrac{1}{2}\\
x = \dfrac{3}{5}
\end{array} \right.\\
Vay\,x = – \dfrac{1}{2};x = 0;x = \dfrac{3}{5}\\
2)\\
A = \dfrac{{{x^2} – 25}}{{{x^2} + 10x + 25}}\\
= \dfrac{{\left( {x – 5} \right)\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}} = \dfrac{{x – 5}}{{x + 5}}\\
B = \dfrac{{{x^3} – 27}}{{{x^3} + 3{x^2} + 9x}}\\
= \dfrac{{\left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{x\left( {{x^2} + 3x + 9} \right)}}\\
= \dfrac{{x – 3}}{x}\\
C = \dfrac{{3{x^2} – 12}}{{6{x^2} + 24x + 24}}\\
= \dfrac{{3\left( {{x^2} – 4} \right)}}{{6\left( {{x^2} + 4x + 4} \right)}}\\
= \dfrac{{3\left( {x – 2} \right)\left( {x + 2} \right)}}{{6.{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{x – 2}}{{2\left( {x + 2} \right)}}\\
= \dfrac{{x – 2}}{{2x + 4}}
\end{array}$