Phân tích đa thức thành nhân tử (2x-3)^3- (5y+2)^3 2x(5x-y)-15x+3y Tìm x 4(x+2)-2x-4 09/07/2021 Bởi Cora Phân tích đa thức thành nhân tử (2x-3)^3- (5y+2)^3 2x(5x-y)-15x+3y Tìm x 4(x+2)-2x-4
Giải thích các bước giải: Ta có: \(\begin{array}{l}{\left( {2x – 3} \right)^3} – {\left( {5y + 2} \right)^3}\\ = \left[ {\left( {2x – 3} \right) – \left( {5y + 2} \right)} \right].\left[ {{{\left( {2x – 3} \right)}^2} + \left( {2x – 3} \right)\left( {5y + 2} \right) + {{\left( {5y + 2} \right)}^2}} \right]\\ = \left( {2x – 5y – 5} \right).\left[ {\left( {4{x^2} – 12x + 9} \right) + \left( {10xy + 4x – 15y – 6} \right) + \left( {25{y^2} + 20y + 4} \right)} \right]\\ = \left( {2x – 5y – 5} \right).\left( {4{x^2} + 10xy + 25{y^2} – 8x + 5y + 7} \right)\\b,\\2x\left( {5x – y} \right) – 15x + 3y\\ = 2x\left( {5x – y} \right) – \left( {15x – 3y} \right)\\ = 2x\left( {5x – y} \right) – 3.\left( {5x – y} \right)\\ = \left( {5x – y} \right)\left( {2x – 3} \right)\\c,\\4\left( {x + 2} \right) – 2x – 4 = 0\\ \Leftrightarrow 4.\left( {x + 2} \right) – 2.\left( {x + 2} \right) = 0\\ \Leftrightarrow 2\left( {x + 2} \right) = 0\\ \Leftrightarrow x + 2 = 0\\ \Leftrightarrow x = – 2\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {2x – 3} \right)^3} – {\left( {5y + 2} \right)^3}\\
= \left[ {\left( {2x – 3} \right) – \left( {5y + 2} \right)} \right].\left[ {{{\left( {2x – 3} \right)}^2} + \left( {2x – 3} \right)\left( {5y + 2} \right) + {{\left( {5y + 2} \right)}^2}} \right]\\
= \left( {2x – 5y – 5} \right).\left[ {\left( {4{x^2} – 12x + 9} \right) + \left( {10xy + 4x – 15y – 6} \right) + \left( {25{y^2} + 20y + 4} \right)} \right]\\
= \left( {2x – 5y – 5} \right).\left( {4{x^2} + 10xy + 25{y^2} – 8x + 5y + 7} \right)\\
b,\\
2x\left( {5x – y} \right) – 15x + 3y\\
= 2x\left( {5x – y} \right) – \left( {15x – 3y} \right)\\
= 2x\left( {5x – y} \right) – 3.\left( {5x – y} \right)\\
= \left( {5x – y} \right)\left( {2x – 3} \right)\\
c,\\
4\left( {x + 2} \right) – 2x – 4 = 0\\
\Leftrightarrow 4.\left( {x + 2} \right) – 2.\left( {x + 2} \right) = 0\\
\Leftrightarrow 2\left( {x + 2} \right) = 0\\
\Leftrightarrow x + 2 = 0\\
\Leftrightarrow x = – 2
\end{array}\)