Please dijawab cepet

Please dijawab cepet

Jawaban Terkonfirmasi

berdasarkan soal diatas maka

k. $frac{sqrt{2}+sqrt{7}}{2-sqrt{3}} =$

= $frac{sqrt{2}+sqrt{7}}{2-sqrt{3}} . frac{2+sqrt{3}}{2+sqrt{3}} = frac{(2+sqrt{3}).(sqrt{2}+sqrt{7})}{4-3}$

= $frac{(2+sqrt{3}).(sqrt{2}+sqrt{7})}{1} = 2sqrt{2}+2sqrt{7}+sqrt{6}+sqrt{21}$

$= 4sqrt{7}+sqrt{6}+sqrt{21}$

l. $frac{2sqrt{5}-3sqrt{2}}{5sqrt{5}+2sqrt{2}} =$

$=frac{2sqrt{5}-3sqrt{2}}{5sqrt{5}+2sqrt{2}} .frac{5sqrt{5}-2sqrt{2}}{5sqrt{5}-2sqrt{2}} = frac{(2sqrt{5}-3sqrt{2})(2sqrt{5}-3sqrt{2})}{25.5 - 4.2}$

$= frac{(2sqrt{5}-3sqrt{2})(2sqrt{5}-3sqrt{2})}{125 - 8} = frac{(2sqrt{5}.2sqrt{5}-2sqrt{5}.3sqrt{2}-3sqrt{2}. 3sqrt{2})}{117}$

$= frac{(4.5-2.3sqrt{10}.-9.4)}{117}$

$=frac{20-6sqrt{10}.-36}{117}$

a. $frac{4-2sqrt{3}}{4+2sqrt{3}} + frac{4+2sqrt{3}}{4-2sqrt{3}} =$

$=frac{4-2sqrt{3}}{4+2sqrt{3}} + frac{4+2sqrt{3}}{4-2sqrt{3}} =frac{(4-2sqrt{3})(4+2sqrt{3}) + (4+2sqrt{3})(4+2sqrt{3})}{(4+2sqrt{3})(4-2sqrt{3})} =$

$=frac{(4-2sqrt{3})(4+2sqrt{3}) + (4+2sqrt{3})(4+2sqrt{3})}{(4+2sqrt{3})(4-2sqrt{3})} =frac{(16-4.3) + (4.4+4.2sqrt{3}+2sqrt{3}.4+2sqrt{3}.2sqrt{3})}{(16-4.3)} =$

$=frac{(16-12) + (16+8sqrt{3}+12sqrt{3}+4.3)}{(16-12)} = frac{(4) + (28+20sqrt{3})}{(4)} = frac{(32+20sqrt{3})}{(4)}$

= $= frac{(32+20sqrt{3})}{(4)}$

$= {(8+5sqrt{3})}$

3. Diketahui :

$p = frac{1-sqrt{2}}{1+sqrt{2}}$

$q= frac{1+sqrt{2}}{1-sqrt{2}}$

Ditanya

a. p+q

b. p²-q

c.p²+q²

d. p²-q²

Dijawab =

a. p+q =

= $frac{1-sqrt{2}}{1+sqrt{2}}$ + $frac{1+sqrt{2}}{1-sqrt{2}}$

$=frac{1-sqrt{2}}{1+sqrt{2}} + frac{1+sqrt{2}}{1-sqrt{2}} =frac{(1-sqrt{2})(1-sqrt{2})+(1+sqrt{2})(1+sqrt{2})}{(1+sqrt{2})(1-sqrt{2})}$

$=frac{(1-sqrt{2})(1-sqrt{2})+(1+sqrt{2})(1+sqrt{2})}{(1+sqrt{2})(1-sqrt{2})}= frac{(1-2)+(1+2sqrt{2}+2)}{1-2}$

$= frac{(1-2)+(1+2sqrt{2}+2)}{1-2} =frac{(-1)+(3+2sqrt{2})}{-1}$

$=frac{(2+2sqrt{2})}{-1}$

$-2-2sqrt{2}$

b. p²-q

= $(frac{1-sqrt{2}}{1+sqrt{2}})^{2} - frac{1+sqrt{2}}{1-sqrt{2}}$

$= (frac{1-sqrt{2}}{1+sqrt{2}}) (frac{1-sqrt{2}}{1+sqrt{2}}) - frac{1+sqrt{2}}{1-sqrt{2}}$

$= (frac{1-sqrt{2}}{1+sqrt{2}}) (frac{1-sqrt{2}}{1+sqrt{2}}) - frac{1+sqrt{2}}{1-sqrt{2}} = frac{1-2}{1+2.2+4} - frac{1+sqrt{2}}{1-sqrt{2}}$

$= frac{-1}{9} - frac{1+sqrt{2}}{1-sqrt{2}}$

$= frac{-1}{9} - frac{1+sqrt{2}}{1-sqrt{2}} = frac{-(1-sqrt{2})-9(1+sqrt{2})}{9(1-sqrt{2})}$

$= frac{-(1-sqrt{2})-9(1+sqrt{2})}{9(1-sqrt{2})} = frac{-1+sqrt{2})-9-9sqrt{2})}{9-9sqrt{2})}$

$= frac{-1+sqrt{2}-9-9sqrt{2}}{9-9sqrt{2}}$

$= frac{-10-8sqrt{2}}{9-9sqrt{2}}$

c. p²+q²

$( frac{1-sqrt{2}}{1+sqrt{2}})^{2} + (frac{1+sqrt{2}}{1-sqrt{2}} )^{2}=$

$=( frac{1-sqrt{2}}{1+sqrt{2}})( frac{1-sqrt{2}}{1+sqrt{2}}) + (frac{1+sqrt{2}}{1-sqrt{2}} )(frac{1+sqrt{2}}{1-sqrt{2}} )$

$=( frac{1-2sqrt{2}+2}{1+2sqrt{2}+2})+ (frac{1+2sqrt{2}+2}{1-2sqrt{2}+2} )$

$=( frac{3-2sqrt{2}}{3+2sqrt{2}})+ (frac{3+2sqrt{2}}{3-2sqrt{2}} )= frac{(3-2sqrt{2})(3-2sqrt{2})+ (3+2sqrt{2})(3+2sqrt{2})}{(3+2sqrt{2})(3-2sqrt{2})}$

$= frac{(3-2sqrt{2})(3-2sqrt{2})+ (3+2sqrt{2})(3+2sqrt{2})}{(3+2sqrt{2})(3-2sqrt{2})} = frac{(9-2.3.sqrt{2}+4.4)+ (9+2.3.sqrt{2}+4.4)}{9-4.2}$

$= frac{(9-2.3.sqrt{2}+4.4)+ (9+2.3.sqrt{2}+4.4)}{9-4.2} =frac{(9-6sqrt{2}+16)+(9+6sqrt{2}+16)}{9-8}$

$=frac{(25-6sqrt{2})+25+6sqrt{2}}{1}$

$=(25-6sqrt{2})+25+6sqrt{2}$

=50

d.p²-q²

$( frac{1-sqrt{2}}{1+sqrt{2}})^{2} - (frac{1+sqrt{2}}{1-sqrt{2}} )^{2}=$

$=( frac{1-sqrt{2}}{1+sqrt{2}})( frac{1-sqrt{2}}{1+sqrt{2}}) - (frac{1+sqrt{2}}{1-sqrt{2}} )(frac{1+sqrt{2}}{1-sqrt{2}} )$

$=( frac{1-2sqrt{2}+2}{1+2sqrt{2}+2})- (frac{1+2sqrt{2}+2}{1-2sqrt{2}+2} )$

$=( frac{3-2sqrt{2}}{3+2sqrt{2}})- (frac{3+2sqrt{2}}{3-2sqrt{2}} )= frac{(3-2sqrt{2})(3-2sqrt{2})- (3+2sqrt{2})(3+2sqrt{2})}{(3+2sqrt{2})(3-2sqrt{2})}$