bang cach rut gon phan thuc chung to moi cap phan thuc sau bang nhau 9x^2-1/12x^2 +4x va 9x^2-6x+1/12x^2-4x 24/08/2021 Bởi Harper bang cach rut gon phan thuc chung to moi cap phan thuc sau bang nhau 9x^2-1/12x^2 +4x va 9x^2-6x+1/12x^2-4x
Đáp án: $\begin{array}{l} + )\frac{{9{x^2} – 1}}{{12{x^2} + 4x}} = \frac{{{{\left( {3x} \right)}^2} – 1}}{{4x\left( {3x + 1} \right)}} = \frac{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}{{4x\left( {3x + 1} \right)}} = \frac{{3x – 1}}{{4x}}\\ + )\frac{{9{x^2} – 6x + 1}}{{12{x^2} – 4x}} = \frac{{{{\left( {3x – 1} \right)}^2}}}{{4x\left( {3x – 1} \right)}} = \frac{{3x – 1}}{{4x}}\\Vậy\,\frac{{9{x^2} – 1}}{{12{x^2} + 4x}} = \frac{{9{x^2} – 6x + 1}}{{12{x^2} – 4x}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
+ )\frac{{9{x^2} – 1}}{{12{x^2} + 4x}} = \frac{{{{\left( {3x} \right)}^2} – 1}}{{4x\left( {3x + 1} \right)}} = \frac{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}{{4x\left( {3x + 1} \right)}} = \frac{{3x – 1}}{{4x}}\\
+ )\frac{{9{x^2} – 6x + 1}}{{12{x^2} – 4x}} = \frac{{{{\left( {3x – 1} \right)}^2}}}{{4x\left( {3x – 1} \right)}} = \frac{{3x – 1}}{{4x}}\\
Vậy\,\frac{{9{x^2} – 1}}{{12{x^2} + 4x}} = \frac{{9{x^2} – 6x + 1}}{{12{x^2} – 4x}}
\end{array}$