Đáp án:
\(\begin{array}{l}
a,\,\,\,\,2.{\left( {x - 7} \right)^2}\\
b,\,\,\,\,5.\left( {x - 2y} \right)\left( {x + 2y} \right)\\
c,\,\,\,\,3.{\left( {a + b} \right)^2}\\
d,\,\,\,\,4.\left( {{x^2} - 3} \right)\left( {{x^2} + 3} \right)\\
e,\,\,\,\,2.\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
f,\,\,\,\,\left( {a - 2y - 4} \right)\left( {a + 2y - 4} \right)\\
g,\,\,\,\,\left( {11 - x - y} \right).\left( {11 + x + y} \right)\\
h,\,\,\,\,\left( {{x^2} + x + 1} \right).\left( {y - x + 1} \right)\\
i,\,\,\,\,4.\left( {5 - y} \right).\left( {25 + 5y + {y^2}} \right)\\
j,\,\,\,\,7.\left( {x + 11} \right).\left( {x - 1} \right)\\
k,\,\,\,\,\left( {4x - 5y - 7} \right).\left( {4x + 5y - 17} \right)\\
l,\,\,\,\,\left( {6 - y - 3x} \right)\left( {6 + y + 3x} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2{x^2} + 98 - 28x\\
= 2.\left( {{x^2} + 49 - 14x} \right)\\
= 2.\left( {{x^2} - 14x + 49} \right)\\
= 2.\left( {{x^2} - 2.x.7 + {7^2}} \right)\\
= 2.{\left( {x - 7} \right)^2}\\
b,\\
5{x^2} - 20{y^2}\\
= 5.\left( {{x^2} - 4{y^2}} \right)\\
= 5.\left[ {{x^2} - {{\left( {2y} \right)}^2}} \right]\\
= 5.\left( {x - 2y} \right)\left( {x + 2y} \right)\\
c,\\
3{a^2} + 6ab + 3{b^2}\\
= 3.\left( {{a^2} + 2ab + {b^2}} \right)\\
= 3.{\left( {a + b} \right)^2}\\
d,\\
4{x^4} - 36\\
= 4.\left( {{x^4} - 9} \right)\\
= 4.\left[ {{{\left( {{x^2}} \right)}^2} - {3^2}} \right]\\
= 4.\left( {{x^2} - 3} \right)\left( {{x^2} + 3} \right)\\
e,\\
2{x^3} + 16\\
= 2.\left( {{x^3} + 8} \right)\\
= 2.\left( {{x^3} + {2^3}} \right)\\
= 2.\left( {x + 2} \right).\left( {{x^2} - x.2 + {2^2}} \right)\\
= 2.\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
f,\\
{\left( {a - 4} \right)^2} - 4{y^2}\\
= {\left( {a - 4} \right)^2} - {\left( {2y} \right)^2}\\
= \left[ {\left( {a - 4} \right) - 2y} \right].\left[ {\left( {a - 4 + 2y} \right)} \right]\\
= \left( {a - 4 - 2y} \right)\left( {a - 4 + 2y} \right)\\
= \left( {a - 2y - 4} \right)\left( {a + 2y - 4} \right)\\
g,\\
121 - \left( {{x^2} + 2xy + {y^2}} \right)\\
= {11^2} - {\left( {x + y} \right)^2}\\
= \left[ {11 - \left( {x + y} \right)} \right].\left[ {11 + \left( {x + y} \right)} \right]\\
= \left( {11 - x - y} \right).\left( {11 + x + y} \right)\\
h,\\
y\left( {{x^2} + x + 1} \right) - \left( {{x^3} - 1} \right)\\
= y\left( {{x^2} + x + 1} \right) - \left( {{x^3} - {1^3}} \right)\\
= y\left( {{x^2} + x + 1} \right) - \left( {x - 1} \right).\left( {{x^2} + x.1 + {1^2}} \right)\\
= y\left( {{x^2} + x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right).\left[ {y - \left( {x - 1} \right)} \right]\\
= \left( {{x^2} + x + 1} \right).\left( {y - x + 1} \right)\\
i,\\
500 - 4{y^3}\\
= 4.\left( {125 - {y^3}} \right)\\
= 4.\left( {{5^3} - {y^3}} \right)\\
= 4.\left( {5 - y} \right).\left( {{5^2} + 5.y + {y^2}} \right)\\
= 4.\left( {5 - y} \right).\left( {25 + 5y + {y^2}} \right)\\
j,\\
4{\left( {2x + 1} \right)^2} - 9.{\left( {x - 3} \right)^2}\\
= {2^2}.{\left( {2x + 1} \right)^2} - {3^2}.{\left( {x - 3} \right)^2}\\
= {\left[ {2.\left( {2x + 1} \right)} \right]^2} - {\left[ {3.\left( {x - 3} \right)} \right]^2}\\
= {\left( {4x + 2} \right)^2} - {\left( {3x - 9} \right)^2}\\
= \left[ {\left( {4x + 2} \right) - \left( {3x - 9} \right)} \right].\left[ {\left( {4x + 2} \right) + \left( {3x - 9} \right)} \right]\\
= \left( {4x + 2 - 3x + 9} \right).\left( {4x + 2 + 3x - 9} \right)\\
= \left( {x + 11} \right).\left( {7x - 7} \right)\\
= 7.\left( {x + 11} \right).\left( {x - 1} \right)\\
k,\\
16.{\left( {x - 3} \right)^2} - 25.{\left( {y - 1} \right)^2}\\
= {4^2}.{\left( {x - 3} \right)^2} - {5^2}.{\left( {y - 1} \right)^2}\\
= {\left[ {4.\left( {x - 3} \right)} \right]^2} - {\left[ {5.\left( {y - 1} \right)} \right]^2}\\
= {\left( {4x - 12} \right)^2} - {\left( {5y - 5} \right)^2}\\
= \left[ {\left( {4x - 12} \right) - \left( {5y - 5} \right)} \right].\left[ {\left( {4x - 12} \right) + \left( {5y - 5} \right)} \right]\\
= \left( {4x - 12 - 5y + 5} \right).\left( {4x - 12 + 5y - 5} \right)\\
= \left( {4x - 5y - 7} \right).\left( {4x + 5y - 17} \right)\\
l,\\
36 - \left( {{y^2} + 6xy + 9{y^2}} \right)\\
= {6^2} - \left[ {{y^2} + 2.y.3x + {{\left( {3x} \right)}^2}} \right]\\
= {6^2} - {\left( {y + 3x} \right)^2}\\
= \left[ {6 - \left( {y + 3x} \right)} \right].\left[ {6 + \left( {y + 3x} \right)} \right]\\
= \left( {6 - y - 3x} \right)\left( {6 + y + 3x} \right)
\end{array}\)
