Let's simplify the following expressions:

$$
\begin{gathered}
\frac{\sqrt{a}+\sqrt{b}-\sqrt{a-b}}{\sqrt{a}+\sqrt{b}+\sqrt{a-b}} \\
\frac{\sqrt{a}-\sqrt{a-1}-\sqrt{a^{2}-a}+\sqrt{a+2-a-1}}{\sqrt{a^{2}-a}+\sqrt{a^{2}-a-1}-\sqrt{a}-\sqrt{a-1}}
\end{gathered}
$$

Calculate their values to three significant figures (2) when $a=3$ and $a=4$.
I. Extend (1) by its numerator, then simplify the expression obtained and divide it by $\sqrt{a}+\sqrt{a-b}$.

$$
\begin{gathered}
K=\frac{2 a+2 \sqrt{a b}-2 \sqrt{a(a-b)}-2 \sqrt{b(a-b)}}{2 b+2 \sqrt{a b}}= \\
=\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{a-b})}{\sqrt{b}(\sqrt{a}+\sqrt{b})}=\frac{\sqrt{a}-\sqrt{a-b}}{\sqrt{b}}=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{a-b}}
\end{gathered}
$$

II. Denote the square roots in the numerator of $L$ as $A, B, C, D$, respectively, write the numerator as the difference of two differences, and transform this and the product of the two differences into a quotient, rationalizing each term of the quotient separately:

$$
\frac{(A-B)-(C-D)}{(A-B)(C-D)}=\frac{1}{C-D}-\frac{1}{A-B}=(C+D)-(A+B)
$$

In the last step, we used the fact that the value of the denominators is $C^{2}-D^{2}=A^{2}-B^{2}=1$. The found quotient is equal to the denominator of $L$, hence

$$
L=(A-B)(C-D)=(\sqrt{a}-\sqrt{a-1})\left(\sqrt{a^{2}-a}-\sqrt{a^{2}-a-1}\right)
$$

(We assume, of course, that there is no negative number under any square root, i.e., $a^{2}-a-1=(a-0.5)^{2}-1.25 \geq 0$, and $a-1 \geq 0$ also holds.)

III. The value of $L$ when $a=3$, calculated to three significant digits:

$$
(\sqrt{3}-\sqrt{2})(\sqrt{6}-\sqrt{5})=\sqrt{18}+\sqrt{10}-\sqrt{15}-\sqrt{12} \approx+0.0678
$$

This is indeed the negative of the $x$ calculated in Exercise 957.1

For $a=4$, the first significant digit of $L$ has a place value of 0.01, as the school four-digit function table data, rounded to three decimal places, gives

$$
\begin{gathered}
L=(2-\sqrt{3})(\sqrt{12}-\sqrt{11})=\sqrt{48}-\sqrt{44}-6+\sqrt{33} \approx \\
\approx 6.928-6.633-6+5.745=0.040
\end{gathered}
$$

Since no rounding error is greater than 0.0005, and the combined error of the three roundings is 0.0015, $0.0385<L<0.0415$. Therefore, the value of $L$ rounded to four decimal places must be given. By rationalizing the numerator,

$$
L=\frac{1}{2+\sqrt{3}} \cdot \frac{1}{\sqrt{12}+\sqrt{11}}=\frac{1}{\sqrt{48}+\sqrt{44}+6+\sqrt{33}} \approx \frac{1}{25.306}
$$

Considering the upper limits of the rounding errors again, the value of $L$ falls between the reciprocals of 25.3075 and 25.3045. Rounding the former, which is the lower limit, downward, and the latter upward, $0.03951<L<0.03952$, so the desired value is 0.0395 (rounded down).

Döme Éva (Makó, József A. g. II. o. t.)[^0]


[^0]:    ${ }^{1}$ K. M. L. 31 (1965) 121. o.