tolong dong jawab yg ini plissss

Mapel : Matematika
Materi : Akar Pangkat

Soal No. 1
[tex]\sqrt{3}\times \sqrt{27}=(\sqrt{3}\times \sqrt{3^3})=\sqrt{3^4}=(3^4)^{ \frac{1}{2} }=3^{(4\times \frac{1}{2})}=3^2=\boxed{9}[/tex]

Soal No. 2
[tex]=\sqrt{2}\times \sqrt{6}\times \sqrt{12} \\ =(\sqrt{2}\times \sqrt{6})\times \sqrt{12} \\= \sqrt{12}\times \sqrt{12}\\= (\sqrt{12})^2\\ =\boxed{12}[/tex]

Soal No. 3
[tex]=(3\sqrt{5}+4\sqrt{2})^2 \\ =(3\sqrt{5})^2+(2\times3\sqrt{5}\times4\sqrt{2})+(4\sqrt{2})^2 \\ =(9\times5)+(24\sqrt{10})+(16\times2) \\ =45+(24\sqrt{10})+32 \\ =\boxed{77+24\sqrt{10}}[/tex]

Soal No. 4
[tex]=(2\sqrt{3}-3\sqrt{7})^2 \\ =(2\sqrt{3})^2-(2\times2\sqrt{3}\times3\sqrt{7})+(3\sqrt{7})^2 \\ =(4\times3)-(12\sqrt{21})+(9\times7) \\ =12-(12\sqrt{21})+63 \\ =\boxed{75-12\sqrt{21}}[/tex]

Soal No. 5
Diketahui :
Panjang = (5 + √2) cm
Lebar = (5 - √2) cm

Ditanya :
Keliling dan Luas Persegi Panjang

Jawab :
a. Keliling Persegi Panjang
[tex]Keliling = 2(P + L) \\ Keliling = 2((5+ \sqrt{2})+(5- \sqrt{2}))\\ Keliling = 2((5+5)+( \sqrt{2}- \sqrt{2}))\\ Keliling = 2(10+0)\\ \boxed{Keliling = 20\ cm}[/tex]

b. Luas Persegi Panjang
[tex]Luas=P\times L \\ Luas = (5+ \sqrt{2})\times(5- \sqrt{2})......Ingat!\ (a+b)(a-b)=a^2-b^2 \\ Luas=5^2-(\sqrt{2})^2 \\ Luas=25-2 \\ \boxed{Luas =23\ cm^2}[/tex]

***Semoga Terbantu***