Question

Va roggggg!
$\frac{ \sqrt{3 }-1 }{ \sqrt{2} }+ \frac{ \sqrt{3}- \sqrt{2} }{ \sqrt{6} } +...+ \frac{ \sqrt{10}- \sqrt{9} }{ \sqrt{90} }$

Answer 1

$\frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ 1 }{ \sqrt{2} }+ \frac{ \sqrt{3} }{ \sqrt{6} }- \frac{ \sqrt{2} }{ \sqrt{6} }+ \frac{ \sqrt{4} }{ \sqrt{12} }- \frac{ \sqrt{3} }{ \sqrt{12} }+.....+ \frac{ \sqrt{10} }{ \sqrt{90} }- \frac{ \sqrt{9} }{ \sqrt{90} }=$
amplificam al doilea termen cu $\sqrt{3} ...... \frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ \sqrt{3} }{ \sqrt{6} }+ \frac{ \sqrt{3} }{ \sqrt{6} }- \frac{ \sqrt{2} }{ \sqrt{6} }+ \frac{ \sqrt{4} }{ \sqrt{12} }- \frac{ \sqrt{3} }{ \sqrt{12} }+.....+ \frac{ \sqrt{10} }{ \sqrt{90} }- \frac{ \sqrt{9} }{ \sqrt{90} }=$
dupa cum se poate observa termenul 2 si 3 sunt egali si cu semn opus deci vor fi 0
$\frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ \sqrt{2} }{ \sqrt{6} }+ \frac{ \sqrt{4} }{ \sqrt{12} }- \frac{ \sqrt{3} }{ \sqrt{12} }+.....+ \frac{ \sqrt{10} }{ \sqrt{90} }- \frac{ \sqrt{9} }{ \sqrt{90} }=$
amplificam al doilea termen cu $\sqrt{2}$
si observam dinnou ca termenii 2 si 3 sunt egali de semn opus
...............
$\frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ \sqrt{10} }{ \sqrt{90} }+ \frac{ \sqrt{10} }{ \sqrt{90} }- \frac{ \sqrt{9} }{ \sqrt{90} }=$
$\frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ \sqrt{9} }{ \sqrt{90} }=\frac{ \sqrt{3*45} }{ \sqrt{90} } - \frac{ \sqrt{9} }{ \sqrt{90} }= \frac{ \sqrt{135} - \sqrt{9} }{ \sqrt{90} } = \frac{3 \sqrt{15}-3 }{ 3\sqrt{10} } =\frac{ \sqrt{15}-1 }{ \sqrt{10} }$