bài 1 : a) x^2y – xy^2 , b)x^2 – y^2 + 5x – 5y bài 2 : a) (a-b)^2 + 4ab , b) 7x^2 + 14x + 7/3x^2 + 3x 27/07/2021 Bởi Bella bài 1 : a) x^2y – xy^2 , b)x^2 – y^2 + 5x – 5y bài 2 : a) (a-b)^2 + 4ab , b) 7x^2 + 14x + 7/3x^2 + 3x
Đáp án: $\begin{array}{l}1)\\a){x^2}y – x{y^2} = xy\left( {x – y} \right)\\b){x^2} – {y^2} + 5x – 5y\\ = \left( {x – y} \right)\left( {x + y} \right) + 5\left( {x – y} \right)\\ = \left( {x – y} \right)\left( {x + y + 5} \right)\\2)\\a){\left( {a – b} \right)^2} + 4ab\\ = {a^2} – 2ab + {b^2} + 4ab\\ = {a^2} + 2ab + {b^2}\\ = {\left( {a + b} \right)^2}\\b)\frac{{7{x^2} + 14x + 7}}{{3{x^2} + 3x}} = \frac{{7\left( {{x^2} + 2x + 1} \right)}}{{3x\left( {x + 1} \right)}}\\ = \frac{{7{{\left( {x + 1} \right)}^2}}}{{3x\left( {x + 1} \right)}} = \frac{{7\left( {x + 1} \right)}}{{3x}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
1)\\
a){x^2}y – x{y^2} = xy\left( {x – y} \right)\\
b){x^2} – {y^2} + 5x – 5y\\
= \left( {x – y} \right)\left( {x + y} \right) + 5\left( {x – y} \right)\\
= \left( {x – y} \right)\left( {x + y + 5} \right)\\
2)\\
a){\left( {a – b} \right)^2} + 4ab\\
= {a^2} – 2ab + {b^2} + 4ab\\
= {a^2} + 2ab + {b^2}\\
= {\left( {a + b} \right)^2}\\
b)\frac{{7{x^2} + 14x + 7}}{{3{x^2} + 3x}} = \frac{{7\left( {{x^2} + 2x + 1} \right)}}{{3x\left( {x + 1} \right)}}\\
= \frac{{7{{\left( {x + 1} \right)}^2}}}{{3x\left( {x + 1} \right)}} = \frac{{7\left( {x + 1} \right)}}{{3x}}
\end{array}$