A)(x+y)^3-x^3-y^3 B) x^5-5x^3+4x C) x^3-11x^2+30x Phân tính da thuc thanh nhan tu 28/09/2021 Bởi Faith A)(x+y)^3-x^3-y^3 B) x^5-5x^3+4x C) x^3-11x^2+30x Phân tính da thuc thanh nhan tu
a) $(x+y)^3 – x^3 – y^3$ $= x^3 + 3x^2 + 3xy^2 + y^3 – x^3 – y^3$ $= 3x^2y + 3xy^2$ $= 3xy(x+y)$ b) $x(x^4 – 5x^2 + 4)$ $= x(x^2-1)(x^2-4)$ $=x(x-1)(x+1)(x-2)(x+2)$ Bình luận
$\begin{array}{l} a)\,\,\,{\left( {x + y} \right)^3} – {x^3} – {y^3}\\ \,\,\,\,\,\, = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} – {x^3} – {y^3}\\ \,\,\,\,\,\, = \,3{x^2}y + 3x{y^2} = 3xy(x + y)\\ b)\,{x^5} – 5{x^3} + 4x\\ \,\,\,\,\, = x\left( {{x^4} – 5{x^2} + 4} \right) = x\left( {{x^4} – 4{x^2} – {x^2} + 4} \right)\\ \,\,\,\, = \,x\left[ {{x^2}\left( {{x^2} – 4} \right) – \left( {{x^2} – 4} \right)} \right] = x\left( {{x^2} – 1} \right)\left( {{x^2} – 4} \right)\\ \,\,\,\,\, = x(x – 1)(x + 1)(x – 2)(x + 2)\\ c)\,\,{x^3} – 11{x^2} + 30x\\ \,\,\,\,\, = \,\,x\left( {{x^2} – 11x + 30} \right) = x\left( {{x^2} – 5x – 6x + 30} \right)\\ \,\,\,\,\, = x\left[ {x(x – 5) – 6(x – 5)} \right]\\ \,\,\,\,\, = x(x – 5)(x – 6) \end{array}$ Bình luận
a) $(x+y)^3 – x^3 – y^3$
$= x^3 + 3x^2 + 3xy^2 + y^3 – x^3 – y^3$
$= 3x^2y + 3xy^2$
$= 3xy(x+y)$
b) $x(x^4 – 5x^2 + 4)$
$= x(x^2-1)(x^2-4)$
$=x(x-1)(x+1)(x-2)(x+2)$
$\begin{array}{l}
a)\,\,\,{\left( {x + y} \right)^3} – {x^3} – {y^3}\\
\,\,\,\,\,\, = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} – {x^3} – {y^3}\\
\,\,\,\,\,\, = \,3{x^2}y + 3x{y^2} = 3xy(x + y)\\
b)\,{x^5} – 5{x^3} + 4x\\
\,\,\,\,\, = x\left( {{x^4} – 5{x^2} + 4} \right) = x\left( {{x^4} – 4{x^2} – {x^2} + 4} \right)\\
\,\,\,\, = \,x\left[ {{x^2}\left( {{x^2} – 4} \right) – \left( {{x^2} – 4} \right)} \right] = x\left( {{x^2} – 1} \right)\left( {{x^2} – 4} \right)\\
\,\,\,\,\, = x(x – 1)(x + 1)(x – 2)(x + 2)\\
c)\,\,{x^3} – 11{x^2} + 30x\\
\,\,\,\,\, = \,\,x\left( {{x^2} – 11x + 30} \right) = x\left( {{x^2} – 5x – 6x + 30} \right)\\
\,\,\,\,\, = x\left[ {x(x – 5) – 6(x – 5)} \right]\\
\,\,\,\,\, = x(x – 5)(x – 6)
\end{array}$