Đáp án+Giải thích các bước giải:
a,
$C=\dfrac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\dfrac{\sqrt{a}+1}{\sqrt{a}+2}+\dfrac{\sqrt{a}-2}{1-\sqrt{a}}(a≥0;a\neq1)$
$C=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{(\sqrt{a}+2)(\sqrt{a}-1)}$
$C=\dfrac{a+3\sqrt{a}+2}{(\sqrt{a}+2)(\sqrt{a}-1)}$
$C=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}$
b,
$a=\sqrt{7-4\sqrt{3}}$
$a=\sqrt{(2-\sqrt{3})^2}$
$a=|2-\sqrt{3}|$
$a=2-\sqrt{3}$
Thay $a$ vào $C$ ta được:
$C=\dfrac{\sqrt{2-\sqrt{3}}+1}{\sqrt{2-\sqrt{3}}-1}$
$=\dfrac{\sqrt{3}-1+\sqrt{2}}{\sqrt{3}-1-\sqrt{2}}$
c,
$C$ nguyên khi $\sqrt{a}+1\vdots \sqrt{a}-1$
$⇔\sqrt{a}-1+2\vdots \sqrt{a}-1$
`⇔2\vdots \sqrt{a}-1⇔\sqrt{a}-1∈Ư(2)={±1;±2}`
$\begin{array}{|c|c|c|c|c|}\hline \sqrt{a}-1&1&-1&2&-2\\\hline a&4&0&9&\emptyset\\\hline \end{array}$
Vậy `a∈{4;0;9}` thì $C$ nguyên
