1. Phân tích bằng cách dùng hằng đẳng thức:
a, $x^{4}$ -1
b, $16x^{4}$ – $81y^{8}$
c, 100x ² – (x ² – 25) ²
2. Phân tích bằng phương pháp nhóm hạng tử và tách hạng tử:
a, x ² – xz -9y ² + 3yz
b, x ³ – x ² -5x+125
c, x ³+2x ²-6x-27
d, x ²(x ² -6)-x ²+6
e, x ²+x-6
f, x ²-3x+2
g, x ²-x-20
h, 2x ²-5x-7
k, x ²-x-12
Đáp án:
b, `x^3 – x^2 – 5x + 125`
`= (x^3 + 125) – (x^2 + 5x)`
`= (x + 5)(x^2 – 5x + 25) – x.(x + 5)`
`=(x + 5)(x^2 – 5x + 25 – x)`
`=(x + 5)(x^2 – 6x + 25)`
c, `x^3 + 2x^2 – 6x – 27`
`=(x^3 – 27) + (2x^2 – 6x)`
`=(x – 3)(x^2 + 3x + 9) – 2x.(x – 3)`
`=(x – 3)(x^2 + 3x + 9 – 2x)`
`=(x – 3)(x^2 + x + 9)`
g, `x^2 – x – 20`
`= x^2 + 4x – 5x – 20`
`= x(x + 4) – 5.(x + 4)`
`=(x – 5)(x + 4)`
h,`2x^2 – 5x – 7`
`= 2x^2 + 2x – 7x – 7`
`= 2x.(x + 1) – 7.(x + 1)`
`=(2x – 7)(x + 1)`
k, `x^2 – x – 12`
`= x^2 + 3x – 4x – 12`
`= x.(x + 3) – 4.(x + 3)`
`=(x – 4)(x + 3)`
Giải thích các bước giải:
Giải thích các bước giải:
$\begin{array}{l}
b){x^3} – {x^2} – 5x + 125\\
= \left( {{x^3} + 5{x^2}} \right) – \left( {6{x^2} + 30x} \right) + \left( {25x + 125} \right)\\
= {x^2}\left( {x + 5} \right) – 6x\left( {x + 5} \right) + 25\left( {x + 5} \right)\\
= \left( {x + 5} \right)\left( {{x^2} – 6x + 25} \right)\\
c){x^3} + 2{x^2} – 6x – 27\\
= \left( {{x^3} – 3{x^2}} \right) + \left( {5{x^2} – 15x} \right) + \left( {9x – 27} \right)\\
= {x^2}\left( {x – 3} \right) + 5x\left( {x – 3} \right) + 9\left( {x – 3} \right)\\
= \left( {x – 3} \right)\left( {{x^2} + 5x + 9} \right)\\
g){x^2} – x – 20\\
= \left( {{x^2} – 5x} \right) + \left( {4x – 20} \right)\\
= \left( {x – 5} \right)x + 4\left( {x – 5} \right)\\
= \left( {x – 5} \right)\left( {x + 4} \right)\\
h)2{x^2} – 5x – 7\\
= \left( {2{x^2} – 7x} \right) + \left( {2x – 7} \right)\\
= \left( {2x – 7} \right)x + \left( {2x – 7} \right)\\
= \left( {2x – 7} \right)\left( {x + 1} \right)\\
k){x^2} – x – 12\\
= \left( {{x^2} – 4x} \right) + \left( {3x – 12} \right)\\
= \left( {x – 4} \right)x + \left( {x – 4} \right).3\\
= \left( {x – 4} \right)\left( {x + 3} \right)
\end{array}$