Toán Lớp 8: Rút gọn phân thức sau
$\frac{x^{3}-4x^{2}+4x-1}{(x-2)(-x+1)+2x(2-x)+x^{2}(x-2)}$
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Toán Lớp 8: Rút gọn phân thức sau
$\frac{x^{3}-4x^{2}+4x-1}{(x-2)(-x+1)+2x(2-x)+x^{2}(x-2)}$
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\dfrac{{{x^3} – 4{x^2} + 4x – 1}}{{\left( {x – 2} \right)\left( { – x + 1} \right) + 2x\left( {2 – x} \right) + {x^2}\left( {x – 2} \right)}}\\
= \dfrac{{\left( {{x^3} – {1^3}} \right) – \left( {4{x^2} – 4x} \right)}}{{\left( {x – 2} \right)\left( {1 – x} \right) – 2x\left( {x – 2} \right) + {x^2}\left( {x – 2} \right)}}\\
= \dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) – 4x\left( {x – 1} \right)}}{{\left( {x – 2} \right)\left( {1 – x – 2x + {x^2}} \right)}}\\
= \dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1 – 4x} \right)}}{{\left( {x – 2} \right)\left( {1 – x – 2x + {x^2}} \right)}}\\
= \dfrac{{\left( {x – 1} \right)\left( {{x^2} – 3x + 1} \right)}}{{\left( {x – 2} \right)\left( {{x^2} – 3x + 1} \right)}}\\
= \dfrac{{x – 1}}{{x – 2}}
\end{array}\]