Toán Lớp 8: B1 phân tích đa thưc thành nhân tử :
x^2y – 2xy^2 + y^3
x^3 + 2 – 2x^2 – x
x^3 + 3x^2 + 3x + 1 – 8z^3
B2 thực hiện rút gọn các biểu thức sau
a, x^2/ x – 3 + 6x / 3- x + 9 / x- 3
b , x-1 / 2x+ 4 . ( 1/x-1 – 1 / x + 1 ) giúp tui vs T_T
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1)a){x^2}y – 2x{y^2} + {y^3}\\
= y\left( {{x^2} – 2xy + {y^2}} \right)\\
= y.{\left( {x – y} \right)^2}\\
b){x^3} + 2 – 2{x^2} – x\\
= {x^3} – x – 2{x^2} + 2\\
= x\left( {{x^2} – 1} \right) – 2\left( {{x^2} – 1} \right)\\
= \left( {{x^2} – 1} \right)\left( {x – 2} \right)\\
= \left( {x + 1} \right)\left( {x – 1} \right)\left( {x – 2} \right)\\
c){x^3} + 3{x^2} + 3x + 1 – 8{z^3}\\
= {\left( {x + 1} \right)^3} – {\left( {2z} \right)^3}\\
= \left( {x + 1 + 2z} \right)\left( {x + 1 – 2z} \right)\\
B2)\\
a)\dfrac{{{x^2}}}{{x – 3}} + \dfrac{{6x}}{{3 – x}} + \dfrac{9}{{x – 3}}\\
= \dfrac{{{x^2} – 6x + 9}}{{x – 3}}\\
= \dfrac{{{{\left( {x – 3} \right)}^2}}}{{x – 3}}\\
= x – 3\\
b)\dfrac{{x – 1}}{{2x + 4}}.\left( {\dfrac{1}{{x – 1}} – \dfrac{1}{{x + 1}}} \right)\\
= \dfrac{{x – 1}}{{2\left( {x + 2} \right)}}.\dfrac{{x + 1 – x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{1}{{2\left( {x + 2} \right)}}.\dfrac{2}{{x + 1}}\\
= \dfrac{1}{{\left( {x + 2} \right)\left( {x + 1} \right)}}
\end{array}$