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wishmaster
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- Oct 11, 2013
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I have a question why is \(\displaystyle \frac{2x^2-1}{x^2+1}\) equal to \(\displaystyle 2-\frac{3}{x^2+1}\) ??
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\begin{aligned}U have a question why is \(\displaystyle \frac{2x^2-1}{x^2+1}\) equal to \(\displaystyle 2-\frac{3}{x^2+1}\) ??
thank you!\begin{aligned}
\frac{2x^2-1}{x^2+1}
&= \frac{2x^2 + 2 - 3}{x^2+1} \\
&= \frac{2(x^2 + 1) - 3}{x^2+1} \\
&= \frac{2(x^2 + 1)}{x^2+1} - \frac 3{x^2+1}\\
&= 2-\frac{3}{x^2+1} \\
\end{aligned}
It is similar to calculating:thank you!
Is this the only way to caclulate it?
[tex]\text{Why is }\,\frac{2x^2-1}{x^2+1}\,\text{ equal to }\,2-\frac{3}{x^2+1}\,?[/tex]
Hello, wishmaster!
Are you familiar with Long Division?
. . [tex]\begin{array}{cccccc}
&&&& 2 \\
&& --&--&-- \\
x^2+1 & | & 2x^2 &-&1 \\
&& 2x^2 &+& 2 \\
&& --&--&-- \\
&&& - & 3 \end{array}[/tex]
Therefore: .[tex]\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}[/tex]