Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\left( {25{x^5} - 5{x^4} + 10{x^2}} \right):5{x^2}\\
= \left( {5{x^3}.5{x^2} - {x^2}.5{x^2} + 2.5{x^2}} \right):5{x^2}\\
= \left[ {5{x^2}.\left( {5{x^3} - {x^2} + 2} \right)} \right]:5{x^2}\\
= 5{x^3} - {x^2} + 2\\
c,\\
\left( {8{x^3} + 1} \right):\left( {4{x^2} - 2x + 1} \right)\\
= \left[ {{{\left( {2x} \right)}^3} + {1^3}} \right]:\left( {4{x^2} - 2x + 1} \right)\\
= \left[ {\left( {2x + 1} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.1 + {1^2}} \right]} \right]:\left( {4{x^2} - 2x + 1} \right)\\
= \left[ {\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right)} \right]:\left( {4{x^2} - 2x + 1} \right)\\
= 2x + 1\\
2,\\
a,\\
\left( {4{x^2} - 9{y^2}} \right):\left( {2x - 3y} \right)\\
= \left[ {{{\left( {2x} \right)}^2} - {{\left( {3y} \right)}^2}} \right]:\left( {2x - 3y} \right)\\
= \left[ {\left( {2x - 3y} \right)\left( {2x + 3y} \right)} \right]:\left( {2x - 3y} \right)\\
= 2x + 3y\\
b,\\
\left( {{x^2} - 3x + xy - 3y} \right):\left( {x + y} \right)\\
= \left[ {\left( {{x^2} + xy} \right) + \left( { - 3x - 3y} \right)} \right]:\left( {x + y} \right)\\
= \left[ {x\left( {x + y} \right) - 3.\left( {x + y} \right)} \right]:\left( {x + y} \right)\\
= \left[ {\left( {x + y} \right)\left( {x - 3} \right)} \right]:\left( {x + y} \right)\\
= x - 3
\end{array}\)
Em xem lại đề câu 1b nhé!
