làm giúp mình Câu 14. Rút gọn các biểu thức sau:,$a)~\sqrt{72}-\sqrt{18}+\sqrt2:$ $b)~\frac{3+\sqrt3}{\sqrt3+1}-\sqrt3$,$c)~P=(\frac2{\sqrt x+1}+\frac2{\sqrt x-1}):\frac4{x-\sqrt x}$ (với $x0,~x
e1)$

Question

làm giúp mình Câu 14. Rút gọn các biểu thức sau:,$a)~\sqrt{72}-\sqrt{18}+\sqrt2:$ $b)~\frac{3+\sqrt3}{\sqrt3+1}-\sqrt3$,$c)~P=(\frac2{\sqrt x+1}+\frac2{\sqrt x-1}):\frac4{x-\sqrt x}$ (với $x>0,~x
e1)$

Accepted Answer

a,
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \sqrt{72} -\sqrt{18} +\sqrt{2}\
=\sqrt{36.2} -\sqrt{9.2} +\sqrt{2}\
=6\sqrt{2} -3\sqrt{2} +\sqrt{2}\
=( 6-3+1)\sqrt{2}\
=4\sqrt{2}
\end{array}$

b,
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \ \ \frac{3+\sqrt{3}}{\sqrt{3} +1} -\sqrt{3}\
\
=\frac{\left(\sqrt{3} ight)^{2} +\sqrt{3}}{\sqrt{3} +1} -\sqrt{3}\
\
=\frac{\sqrt{3}\left(\sqrt{3} +1 ight)}{\sqrt{3} +1} -\sqrt{3}\
\
=\sqrt{3} -\sqrt{3}\
\
=\ 0
\end{array}$

c,
$\displaystyle \begin{array}{{>{\displaystyle}l}}
P=\left(\frac{2}{\sqrt{x} +1} +\frac{2}{\sqrt{x} -1} ight) :\frac{4}{x-\sqrt{x}}\
\
=\left(\frac{2\left(\sqrt{x} -1 ight)}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} +\frac{2\left(\sqrt{x} +1 ight)}{\left(\sqrt{x} -1 ight)\left(\sqrt{x} +1 ight)} ight) :\frac{4}{x-\sqrt{x}}\
\
\
=\frac{2\left(\sqrt{x} -1 ight) +2\left(\sqrt{x} +1 ight)}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} :\frac{4}{x-\sqrt{x}}\
\
=\frac{4\sqrt{x}}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} :\frac{4}{x-\sqrt{x}}\
\
=\frac{4\sqrt{x}}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} .\frac{x-\sqrt{x}}{4}\
\
=\frac{4\sqrt{x}}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} .\frac{\sqrt{x}\left( 1-\sqrt{x} ight)}{4}\
\
=\frac{4\sqrt{x}}{\left(\sqrt{x} +1 ight)\left(\sqrt{x} -1 ight)} .\frac{-\sqrt{x}\left(\sqrt{x} -1 ight)}{4}\
\
=\frac{-\sqrt{x} .\sqrt{x}}{\sqrt{x} +1}\
\
=\frac{-x}{\sqrt{x} +1}\
\end{array}$