Đáp án:
4) 1
Giải thích các bước giải:
\(\begin{array}{l}
1)\dfrac{{\sqrt 2 \left( {\sqrt 3 + \sqrt 7 } \right)}}{{2\sqrt 3 + 2\sqrt 7 }} = \dfrac{{\sqrt 2 \left( {\sqrt 3 + \sqrt 7 } \right)}}{{2\left( {\sqrt 3 + \sqrt 7 } \right)}}\\
= \dfrac{{\sqrt 2 }}{2}\\
2)DK:a \ge 0;b > 0\\
\dfrac{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}{{\sqrt b \left( {\sqrt b + \sqrt a } \right)}} = \sqrt {\dfrac{a}{b}} \\
3)DK:x \ge 0;x \ne 1\\
\dfrac{{x - \sqrt x + 3\sqrt x - 3}}{{\sqrt x - 1}} = \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 3\left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}{{\sqrt x - 1}} = \sqrt x + 3\\
4)DK:x \ge 3\\
\sqrt {x - 3 + 2\sqrt {x - 3} .1 + 1} - \sqrt {x - 3} \\
= \sqrt {{{\left( {\sqrt {x - 3} + 1} \right)}^2}} - \sqrt {x - 3} \\
= \sqrt {x - 3} + 1 - \sqrt {x - 3} = 1
\end{array}\)
