p) x^2 + x – 6
q) x^2 + x – 6
s) x^2 – 5x – 6
t) x^2 – 8x – 9
u x^2 +3x – 18
v) x^2 – 7x +6
w) 3x^2 -7x +2
y) x^4 + 64
p) x^2 + x – 6
q) x^2 + x – 6
s) x^2 – 5x – 6
t) x^2 – 8x – 9
u x^2 +3x – 18
v) x^2 – 7x +6
w) 3x^2 -7x +2
y) x^4 + 64
Từ đầu đến cuối dùng phương pháp nhóm hạng tử trừ câu y dùng phương pháp hằng đăng thức
`p)=x^2+3x-2x`
`=(x+3)(x-2)`
`q) =x^2-3x+2x-6`
`=(x-3)(x+2)`
`s) =x^2-6x+x-6`
`=(x-6)(x+1)`
`t) x^2-9x+x-9`
`=(x-9)(x+1)`
`u) =x^2+6x-3x-18`
`=(x+6)(x-3)`
`v)=x^2-x-6x+6`
`=(x-1)(x-6)`
`w) =3x^2-6x-x+2`
`=(x-2)(3x-1)`
`y)=(x^2)^2 +2.x^2 .8+8^2-2.x^2 .8`
`=(x^2+8)^2-16x^2`
`=(x^2+8)^2-(4x)^2`
`=(x^2-4x+8)(x^2+4x+8)`
Giải thích các bước giải:
$\begin{array}{l}
p){x^2} + x – 6\\
= {x^2} + 3x – 2x – 6\\
= \left( {x + 3} \right)\left( {x – 2} \right)\\
q){x^2} – x – 6\\
= {x^2} – 3x + 2x – 6\\
= \left( {x – 3} \right)\left( {x + 2} \right)\\
s){x^2} – 5x – 6\\
= {x^2} – 6x + x – 6\\
= \left( {x – 6} \right)\left( {x + 1} \right)\\
t){x^2} – 8x – 9\\
= {x^2} – 9x + x – 9\\
= \left( {x – 9} \right)\left( {x + 1} \right)\\
u){x^2} + 3x – 18\\
= {x^2} + 6x – 3x – 18\\
= \left( {x + 6} \right)\left( {x – 3} \right)\\
v){x^2} – 7x + 6\\
= {x^2} – x – 6x + 6\\
= \left( {x – 1} \right)\left( {x – 6} \right)\\
w)3{x^2} – 7x + 2\\
= 3{x^2} – 6x – x + 2\\
= \left( {x – 2} \right)\left( {3x – 1} \right)\\
y){x^4} + 64\\
= {\left( {{x^2}} \right)^2} + 2.{x^2}.8 + {8^2} – 2.{x^2}.8\\
= {\left( {{x^2} + 8} \right)^2} – 16{x^2}\\
= {\left( {{x^2} + 8} \right)^2} – {\left( {4x} \right)^2}\\
= \left( {{x^2} – 4x + 8} \right)\left( {{x^2} + 4x + 8} \right)
\end{array}$