Phân tích thành nhân tử:
a, ( ab – 1 )2 + ( a+b )2
c,x3 – 4×2 +12 x – 27
b, x3 +2×2 +2x + 1
d, x4 – 2×3 +2x +1
e,x4 +2×3 + 2×2 +2x +1
Phân tích thành nhân tử:
a, ( ab – 1 )2 + ( a+b )2
c,x3 – 4×2 +12 x – 27
b, x3 +2×2 +2x + 1
d, x4 – 2×3 +2x +1
e,x4 +2×3 + 2×2 +2x +1
Đáp án:
$\begin{array}{l}
a){\left( {ab – 1} \right)^2} + {\left( {a + b} \right)^2}\\
= {a^2}{b^2} – 2ab + 1 + {a^2} + 2ab + {b^2}\\
= {a^2}{b^2} + 1 + {a^2} + {b^2}\\
= {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\
= \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\
c){x^3} – 4{x^2} + 12x – 27\\
= {x^3} – 27 + \left( { – 4{x^2} + 12x} \right)\\
= \left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right) – 4x\left( {x – 3} \right)\\
= \left( {x – 3} \right)\left( {{x^2} + 3x + 9 – 4x} \right)\\
= \left( {x – 3} \right)\left( {{x^2} – x + 9} \right)\\
b){x^3} + 2{x^2} + 2x + 1\\
= {x^3} + 2{x^2} + x + x + 1\\
= x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\
= x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\
d){x^4} – 2{x^3} + 2x – 1\\
= {x^4} – 2{x^3} + {x^2} – {x^2} + 2x – 1\\
= {x^2}\left( {{x^2} – 2x + 1} \right) – \left( {{x^2} – 2x + 1} \right)\\
= \left( {{x^2} – 2x + 1} \right)\left( {{x^2} – 1} \right)\\
= {\left( {x – 1} \right)^2}\left( {x – 1} \right)\left( {x + 1} \right)\\
= {\left( {x – 1} \right)^3}\left( {x + 1} \right)\\
e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
= {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\
= {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\
= \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\
= {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right)
\end{array}$
Đáp án:
a)(ab−1)2+(a+b)2
=a2b2−2ab+1+a2+2ab+b2
=a2b2+1+a2+b2=a2(b2+1)+(b2+1) = (a2+1)(b2+1)
c)x3−4x2+12x−27
=x3−27+(−4x2+12x)
=(x−3)(x2+3x+9)−4x(x−3)
=(x−3)(x2+3x+9−4x)
=(x−3)(x2−x+9)
b)x3+2x2+2x+1
=x3+2x2+x+x+1
=x(x2+2x+1)+(x+1)
=x(x+1)2+(x+1)
=(x+1)(x(x+1)+1)
=(x+1)(x2+x+1)
d)x4−2x3+2x−1
=x4−2x3+x2−x2+2x−1
=x2(x2−2x+1)−(x2−2x+1)
=(x2−2x+1)(x2−1)
=(x−1)2(x−1)(x+1)
=(x−1)3(x+1)
e)x4+2x3+2x2+2x+1
=x4+2x3+x2+x2+2x+1
=x2(x2+2x+1)+(x2+2x+1)
=(x2+2x+1)(x2+1)
=(x+1)2(x2+1)