發新話題
打印

根號數的逼近

回復 1# Wuwen 的帖子

\(\begin{align}
  & \frac{b}{a}-\frac{b+7a}{a+b} \\
& =\frac{\left( b+\sqrt{7}a \right)\left( b-\sqrt{7}a \right)}{a\left( a+b \right)} \\
& \left( 1 \right)\ b>\sqrt{7}a\quad ,\quad \frac{b}{a}>\frac{b+7a}{a+b} \\
& \left( \frac{b}{a}-\sqrt{7} \right)-\left( \sqrt{7}-\frac{b+7a}{a+b} \right) \\
& =\frac{b}{a}+\frac{b+7a}{a+b}-2\sqrt{7} \\
& =\frac{\left( 7-2\sqrt{7} \right){{a}^{2}}+\left( 2-2\sqrt{7} \right)ab+{{b}^{2}}}{a\left( a+b \right)} \\
& =\frac{\left( b-\sqrt{7}a \right)\left[ b+\left( 2-\sqrt{7} \right)a \right]}{a\left( a+b \right)} \\
& =\frac{{{\left( b-\sqrt{7}a \right)}^{2}}+2a\left( b-\sqrt{7}a \right)}{a\left( a+b \right)}>0 \\
& \frac{b}{a}-\sqrt{7}>\sqrt{7}-\frac{b+7a}{a+b} \\
& \left( 2 \right)\ b<\sqrt{7}a\quad ,\quad \frac{b}{a}<\frac{b+7a}{a+b} \\
& \cdots \cdots  \\
\end{align}\)
同理可證

TOP

發新話題