Tolong yaa kerjain yg dilingkari, sebisanya aja gakpapa. pleasee.... Question from @zaraaaa

October 2018 1 22 Report

Tolong yaa kerjain yg dilingkari, sebisanya aja gakpapa. pleasee..

whongaliem 1a) $\frac{5}{ \sqrt{15} }$ = $\frac{5}{ \sqrt{15} }$ X $\frac{ \sqrt{15} }{ \sqrt{15} }$
= $\frac{5. \sqrt{15} }{15}$
= $\frac{1}{3} \sqrt{15}$
b) dan c) kerjakan seperit di atas sebagaitihan
d) $\frac{ \sqrt{6} }{ \sqrt{24} }$ = $\frac{ \sqrt{6} }{ \sqrt{24} }$ X $\frac{ \sqrt{24} }{ \sqrt{24} }$
= $\frac{ \sqrt{144} }{24}$
= $\frac{12}{24}$
= $\frac{1}{2}$
2b) $\frac{4 - \sqrt{2} }{4 + \sqrt{2} }$ = $\frac{4 - \sqrt{2} }{4 + \sqrt{2} }$ X $\frac{4 - \sqrt{2} }{4 - \sqrt{2} }$
= $\frac{ (4 - \sqrt{2}) ^{2} }{16 - 2}$
= $\frac{16 - 8. \sqrt{2} + 2}{14}$
= $\frac{18 - 8. \sqrt{2} }{14}$
= $\frac{9 - 4. \sqrt{2} }{7}$
= $\frac{1}{7}$ (9 - 4.√2)
d) $\frac{ \sqrt{3} }{ \sqrt{5} - \sqrt{10} }$ = $\frac{ \sqrt{3} }{ \sqrt{5} - \sqrt{10} }$ X $\frac{ \sqrt{5} + \sqrt{10} }{ \sqrt{5} + \sqrt{10} }$
= $\frac{ \sqrt{15} + \sqrt{30} }{5 - 10}$
= $-\frac{1}{5}$ (√15 + √30)
e) $\frac{x.y}{ \sqrt{x} + \sqrt{y} }$ = $\frac{x.y}{ \sqrt{x} + \sqrt{y} }$ X $\frac{ \sqrt{x} - \sqrt{y} }{ \sqrt{x} - \sqrt{y} }$
= $\frac{x.y ( \sqrt{x} - \sqrt{y}) }{x - y}$
= $\frac{x.y}{x - y}$ (√x - √y)
5b) $\sqrt{5 + 2 \sqrt{6} }$ = $\sqrt{3 + 2 \sqrt{6} + 2}$
= $\sqrt{ ( \sqrt{3} + \sqrt{2} )^{2} }$
= √3 + √2
d) $\sqrt{21 - 4 \sqrt{5} }$ = $\sqrt{21 - 2 \sqrt{5 . 2^{2} } }$
= $\sqrt{20 - 2 \sqrt{20} + 1}$
= $\sqrt{ ( \sqrt{20} - \sqrt{1} )^{2} }$
= √20 - √1
= 2√5 - 1 atau - 1 + 2√5
e) $\sqrt{18 + 8 \sqrt{2} }$ + $\sqrt{11 - 6 \sqrt{2} }$ = $\sqrt{18 + 2 \sqrt{2 . 4^{2} } }$ + $\sqrt{11 - 2 \sqrt{2 . 3^{2} } }$
= $\sqrt{16 - 2 \sqrt{32} + 2}$ + $\sqrt{9 - 2 \sqrt{18} + 2}$
= $\sqrt{ ( \sqrt{16} + \sqrt{2} )^{2} }$ + $\sqrt{ ( \sqrt{9} - \sqrt{2}) ^{2} }$
= √16 + √2 + √9 - √2
= √16 + √9
= 4 + 3
= 7

1 votes Thanks 1

zaraaaa Terimakasih bnyak