Serius ya jawabnya.. hitunglah p+q, p-q, dan pxq. serta sederhanakan hasilnya jika a. p = [tex] \frac{3}{ \sqrt{2} -3} [/tex] dan q = [tex] \frac{2}{ \sqrt{2

Soal a.
"p + q"

[tex]=\frac{3}{ \sqrt{2}-3}+ \frac{2}{ \sqrt{2}+3}\\ \\=\frac{3(\sqrt{2}+3)+2(\sqrt{2}-3)}{ (\sqrt{2}-3)(\sqrt{2}+3)}\\ \\=\frac{(3\sqrt{2})+9+(2\sqrt{2})-6}{ 2-9} \\ \\ = \frac{5 \sqrt{2}+3 }{-7} \\ \\ =- \frac{5}{7} \sqrt{2}- \frac{3}{7} [/tex]

"p - q"

[tex]=\frac{3}{ \sqrt{2}-3}- \frac{2}{ \sqrt{2}+3}\\ \\=\frac{3(\sqrt{2}+3)-2(\sqrt{2}-3)}{ (\sqrt{2}-3)(\sqrt{2}+3)}\\ \\=\frac{(3\sqrt{2})+9-(2\sqrt{2})+6}{ 2-9} \\ \\ = \frac{ \sqrt{2}+15 }{-7} \\ \\ =- \frac{1}{7} \sqrt{2}- \frac{15}{7}[/tex]

"p × q"

[tex]=(\frac{3}{ \sqrt{2}-3}) \times (\frac{2}{ \sqrt{2}+3}) \\ \\ = \frac{3\times2}{(\sqrt{2}-3)(\sqrt{2}+3)} \\ \\ =\frac{6}{2-9} \\ \\ =\frac{6}{(-7)} \\ \\ =- \frac{6}{7} [/tex]

Soal b.
"p + q"

[tex]=\frac{ \sqrt{12} }{ \sqrt{11}-3}+ \frac{ \sqrt{3} }{ \sqrt{11}+3}\\ \\=\frac{ 2\sqrt{3} }{ \sqrt{11}-3}+ \frac{ \sqrt{3} }{ \sqrt{11}+3}\\ \\=\frac{ 2\sqrt{3} (\sqrt{11}+3)+ \sqrt{3} (\sqrt{11}-3)}{ (\sqrt{11}-3)(\sqrt{11}+3)}\\ \\=\frac{(2\sqrt{33})+(6 \sqrt{3})+(\sqrt{33})-(3 \sqrt{3}) }{ 11-9} \\ \\ = \frac{3 \sqrt{33}-3 \sqrt{3}}{2} \\ \\ = \frac{3}{2} \sqrt{33}- \frac{3}{2} \sqrt{3} \\ \\=\frac{3}{2} \sqrt{3}\ ( \sqrt{11}-1) [/tex]

"p - q"

[tex]=\frac{ \sqrt{12} }{ \sqrt{11}-3}- \frac{ \sqrt{3} }{ \sqrt{11}+3}\\ \\=\frac{ 2\sqrt{3} }{ \sqrt{11}-3}- \frac{ \sqrt{3} }{ \sqrt{11}+3}\\ \\=\frac{ 2\sqrt{3} (\sqrt{11}+3)- \sqrt{3} (\sqrt{11}-3)}{ (\sqrt{11}-3)(\sqrt{11}+3)}\\ \\=\frac{(2\sqrt{33})+(6 \sqrt{3})-(\sqrt{33})+(3 \sqrt{3}) }{ 11-9} \\ \\ = \frac{\sqrt{33}+9 \sqrt{3}}{2} \\ \\ = \frac{1}{2} \sqrt{33}+ \frac{9}{2} \sqrt{3} \\ \\=\frac{1}{2} \sqrt{3}\ ( \sqrt{11}+9)[/tex]

"p × q"

[tex]=\frac{ \sqrt{12} }{ \sqrt{11}-3}\times \frac{ \sqrt{3} }{ \sqrt{11}+3}\\ \\=\frac{ 2\sqrt{3} }{ \sqrt{11}-3}\times \frac{ \sqrt{3} }{ \sqrt{11}+3} \\ \\ = \frac{2\times( \sqrt{3})^2 }{(\sqrt{11}-3)(\sqrt{11}+3)} \\ \\ =\frac{2\times3}{11-9} \\ \\ =\frac{6}{2} \\ \\ =3 [/tex]

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