Đáp án:
\(\begin{array}{l}
1){\left( {x - 3} \right)^2}\\
2){\left( {5 + x} \right)^2}\\
3){\left( {\dfrac{1}{2}a + 2{b^2}} \right)^2}\\
4){\left( {\dfrac{1}{3} + {y^4}} \right)^2}\\
5)\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\
6)\left( {2y - 5} \right)\left( {4{y^2} + 20y + 25} \right)\\
7)\left( {{a^2} - b} \right)\left( {{a^4} + {a^2}b + {b^2}} \right)\\
8){\left( {x - 5} \right)^2}\\
9)\left( {2x - \dfrac{1}{2}} \right)\left( {4{x^2} + x + \dfrac{1}{4}} \right)\\
10){\left( {x + 2y} \right)^2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1){x^2} - 2.3.x + {3^2} = {\left( {x - 3} \right)^2}\\
2){5^2} + 2.5.x + {x^2} = {\left( {5 + x} \right)^2}\\
3)\dfrac{1}{4}{a^2} + 2a{b^2} + 4{b^4}\\
= {\left( {\dfrac{1}{2}a} \right)^2} + 2.\dfrac{1}{2}a.2{b^2} + {\left( {2{b^2}} \right)^2}\\
= {\left( {\dfrac{1}{2}a + 2{b^2}} \right)^2}\\
4){\left( {\dfrac{1}{3}} \right)^2} - 2.\dfrac{1}{3}.{y^4} + {\left( {{y^4}} \right)^2}\\
= {\left( {\dfrac{1}{3} + {y^4}} \right)^2}\\
5){x^3} + {\left( {2y} \right)^3}\\
= \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\
6){\left( {2y} \right)^3} - {5^3}\\
= \left( {2y - 5} \right)\left( {4{y^2} + 20y + 25} \right)\\
7){\left( {{a^2}} \right)^3} - {b^3}\\
= \left( {{a^2} - b} \right)\left( {{a^4} + {a^2}b + {b^2}} \right)\\
8){x^2} - 2.5.x + {5^2}\\
= {\left( {x - 5} \right)^2}\\
9){\left( {2x} \right)^3} - {\left( {\dfrac{1}{2}} \right)^3}\\
= \left( {2x - \dfrac{1}{2}} \right)\left( {4{x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4}} \right)\\
= \left( {2x - \dfrac{1}{2}} \right)\left( {4{x^2} + x + \dfrac{1}{4}} \right)\\
10){x^2} + 2.x.2y + {\left( {2y} \right)^2}\\
= {\left( {x + 2y} \right)^2}
\end{array}\)
