x mũ 3-x mũ 2+8x-4 4x mux2-25-(2x-5)(2x+7) x mũ 3+27+(x+3)(x-9) 12/07/2021 Bởi Clara x mũ 3-x mũ 2+8x-4 4x mux2-25-(2x-5)(2x+7) x mũ 3+27+(x+3)(x-9)
Giải thích các bước giải: Ta có: \(\begin{array}{l}*)\\4{x^2} – 25 – \left( {2x – 5} \right)\left( {2x + 7} \right)\\ = \left( {4{x^2} – 25} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right)\\ = \left[ {{{\left( {2x} \right)}^2} – {5^2}} \right] – \left( {2x – 5} \right)\left( {2x + 7} \right)\\ = \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right)\\ = \left( {2x – 5} \right).\left[ {\left( {2x + 5} \right) – \left( {2x + 7} \right)} \right]\\ = \left( {2x – 5} \right).\left( { – 2} \right)\\ = – 2.\left( {2x – 5} \right)\\*)\\{x^3} + 27 + \left( {x + 3} \right)\left( {x – 9} \right)\\ = \left( {{x^3} + {3^3}} \right) + \left( {x + 3} \right)\left( {x – 9} \right)\\ = \left( {x + 3} \right)\left( {{x^2} – 3.x + {3^2}} \right) + \left( {x + 3} \right)\left( {x – 9} \right)\\ = \left( {x + 3} \right).\left[ {\left( {{x^2} – 3x + 9} \right) + \left( {x – 9} \right)} \right]\\ = \left( {x + 3} \right).\left( {{x^2} – 2x} \right)\\ = x.\left( {x + 3} \right)\left( {x – 2} \right)\end{array}\) Em xem lại đề câu đầu nhé! Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
4{x^2} – 25 – \left( {2x – 5} \right)\left( {2x + 7} \right)\\
= \left( {4{x^2} – 25} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right)\\
= \left[ {{{\left( {2x} \right)}^2} – {5^2}} \right] – \left( {2x – 5} \right)\left( {2x + 7} \right)\\
= \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right)\\
= \left( {2x – 5} \right).\left[ {\left( {2x + 5} \right) – \left( {2x + 7} \right)} \right]\\
= \left( {2x – 5} \right).\left( { – 2} \right)\\
= – 2.\left( {2x – 5} \right)\\
*)\\
{x^3} + 27 + \left( {x + 3} \right)\left( {x – 9} \right)\\
= \left( {{x^3} + {3^3}} \right) + \left( {x + 3} \right)\left( {x – 9} \right)\\
= \left( {x + 3} \right)\left( {{x^2} – 3.x + {3^2}} \right) + \left( {x + 3} \right)\left( {x – 9} \right)\\
= \left( {x + 3} \right).\left[ {\left( {{x^2} – 3x + 9} \right) + \left( {x – 9} \right)} \right]\\
= \left( {x + 3} \right).\left( {{x^2} – 2x} \right)\\
= x.\left( {x + 3} \right)\left( {x – 2} \right)
\end{array}\)
Em xem lại đề câu đầu nhé!