sederhanakan:a.(2√3+√5)(2√3-√5) b.(√7+√6)

Mapel : Matematika
Materi : Akar Pangkat

Penyederhanaan
Soal a

[tex](2 \sqrt{3}+ \sqrt{5})(2 \sqrt{3}- \sqrt{5}) \\ \\ \text{Ingat bahwa}\ (a+b)(a-b)=a^2-b^2,\ dimana : \\ a=2 \sqrt{3}\ dan\ b=\sqrt{5} \\ \\ =(2 \sqrt{3})^2-(\sqrt{5})^2 \\ \\ =(2^2\times( \sqrt{3})^2)-(\sqrt{5})^2\\ \\ =(4\times3)-5\\ \\ =12-5\\\\=7 [/tex]

Soal b (sepertinya soalnya ada yang kurang. jadi saya coba berikan kemungkinan-kemungkinan soalnya)

Jika :
[tex]=( \sqrt{7}+ \sqrt{6})( \sqrt{7}- \sqrt{6}),\ maka : \\ \\ =( \sqrt{7})^2- (\sqrt{6})^2 \\ \\ =7-6 \\ \\ =1[/tex]

Jika :
[tex]=( \sqrt{7}+ \sqrt{6})( \sqrt{7}+ \sqrt{6}),\ maka : \\ \\ =( \sqrt{7})^2+( \sqrt{7})(\sqrt{6})+(\sqrt{6})( \sqrt{7})+(\sqrt{6})^2 \\ \\ =7+ \sqrt{42}+ \sqrt{42}+ 6 \\ \\ =13+2\sqrt{42}[/tex]

***Semoga Terbantu***

Tambahan :

[tex]=\sqrt{24} + \sqrt{54}- \sqrt{150} \\\\=\sqrt{(4\times6)}+\sqrt{(9\times6)}- \sqrt{(25\times6)}\\\\=2 \sqrt{6}+3 \sqrt{6}-5 \sqrt{6}\\\\=(2+3-5) \sqrt{6}\\\\=(0)\times \sqrt{6}\\\\=0 [/tex]

[tex](12)^{\frac{3}{4}}................Ingat,\ bahwa\ a^{ \frac{m}{n}}= \sqrt[n]{a^m} \\ \\(12)^{\frac{3}{4}}= \sqrt[4]{12^3} [/tex]