phân tích đa thức thành nhân tử
`g)x(x-y)^2+y(x-y)^2 -xy+x^2`
`h)(8-2x^2)^2-18(x+2)(x-2)`
`i) x^3-16x-15x(x-4)`
`k)x^3-7x-6`
phân tích đa thức thành nhân tử
`g)x(x-y)^2+y(x-y)^2 -xy+x^2`
`h)(8-2x^2)^2-18(x+2)(x-2)`
`i) x^3-16x-15x(x-4)`
`k)x^3-7x-6`
Đáp án:
$\begin{array}{l}
g)\left( {x – y} \right)\left( {{x^2} – {y^2} + x} \right)\\
h)2\left( {x – 2} \right)\left( {x + 2} \right)\left( {2{x^2} – 17} \right)\\
i)x\left( {x – 4} \right)\left( {x – 11} \right)\\
k)\left( {x + 1} \right)\left( {x – 3} \right)\left( {x + 2} \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
g)x{\left( {x – y} \right)^2} + y{\left( {x – y} \right)^2} – xy + {x^2}\\
= \left( {x – y} \right)\left[ {x\left( {x – y} \right) + y\left( {x – y} \right) + x} \right]\\
= \left( {x – y} \right)\left[ {\left( {x + y} \right)\left( {x – y} \right) + x} \right]\\
= \left( {x – y} \right)\left( {{x^2} – {y^2} + x} \right)\\
h){\left( {8 – 2{x^2}} \right)^2} – 18\left( {x + 2} \right)\left( {x – 2} \right)\\
= {\left( {2\left( {4 – {x^2}} \right)} \right)^2} – 18\left( {{x^2} – 4} \right)\\
= 4{\left( {{x^2} – 4} \right)^2} – 18\left( {{x^2} – 4} \right)\\
= \left( {{x^2} – 4} \right)\left( {4\left( {{x^2} – 4} \right) – 18} \right)\\
= 2\left( {x – 2} \right)\left( {x + 2} \right)\left( {2{x^2} – 17} \right)\\
i){x^3} – 16x – 15x\left( {x – 4} \right)\\
= \left( {{x^3} – 16x} \right) – 15x\left( {x – 4} \right)\\
= x\left( {{x^2} – 16} \right) – 15x\left( {x – 4} \right)\\
= x\left( {x – 4} \right)\left( {x + 4} \right) – 15x\left( {x – 4} \right)\\
= x\left( {x – 4} \right)\left[ {\left( {x + 4} \right) – 15} \right]\\
= x\left( {x – 4} \right)\left( {x – 11} \right)\\
k){x^3} – 7x – 6\\
= {x^3} + {x^2} – {x^2} – x – 6x – 6\\
= \left( {x + 1} \right)\left( {{x^2} – x – 6} \right)\\
= \left( {x + 1} \right)\left( {{x^2} – 3x + 2x – 6} \right)\\
= \left( {x + 1} \right)\left( {x – 3} \right)\left( {x + 2} \right)
\end{array}$