0 bình luận về “rút gọn biểu thức (x^2+căn (x))/(x+ căn (x)+1)”
Đáp án:
$\begin{array}{l} \dfrac{{{x^2} + \sqrt x }}{{x + \sqrt x + 1}}\\ = \dfrac{{\sqrt x \left( {x\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\ = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\ = \sqrt x .\left( {\sqrt x + 1} \right)\\ = x + \sqrt x \end{array}$
Đáp án:
$\begin{array}{l}
\dfrac{{{x^2} + \sqrt x }}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {x\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \sqrt x .\left( {\sqrt x + 1} \right)\\
= x + \sqrt x
\end{array}$
Ta có :
$\frac{x^2+\sqrt{x}}{x+\sqrt{x}+1}$=$\frac{\sqrt{x}(x\sqrt{x}+1)}{x+\sqrt{x}+1}$ =$\frac{\sqrt{x}(\sqrt{x}+1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}$
⇔$\sqrt{x}$.$(\sqrt{x}$+$1$)
⇒$x$+$\sqrt{x}$
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