Bidang studi:
MatematikaPenulis:
josiahgutierrezDibuat:
1 tahun lalu[tex]= \frac{2}{3+\sqrt2}\\=\frac{2}{3+\sqrt2}\times\frac{3-\sqrt2}{3-\sqrt2}\\=\frac{6-2\sqrt2}{(3)^2-(\sqrt2)^2}\\=\frac{6-2\sqrt2}{9-2}\\=\frac{6-2\sqrt2}{7}\\[/tex]
Penulis:
saraizn7y
Nilai jawaban:
3Jawab:
[tex]\frac{6-2\sqrt{2} )}{7 }[/tex]
Penjelasan dengan langkah-langkah:
1) [tex]\frac{2}{3+\sqrt{2} }[/tex]= [tex]\frac{2}{3+\sqrt{2} }x\frac{{3-\sqrt{2} }}{{3-\sqrt{2} }}[/tex]
= [tex]\frac{2(3-\sqrt{2} )}{(3+\sqrt{2})(3-\sqrt{2})}[/tex]
= [tex]\frac{(2x3)-(2\sqrt{2}) )}{(3+\sqrt{2})(3-\sqrt{2})}[/tex]
= [tex]\frac{6-2\sqrt{2} )}{(3+\sqrt{2})(3-\sqrt{2})}[/tex] [tex](Gunakan Rumus (a-b)(a+b)=a^{2} - b^{2} )[/tex]
= [tex]\frac{6-2\sqrt{2} )}{3^{2} x\sqrt{2}^{2} }[/tex]
= [tex]\frac{6-2\sqrt{2} )}{9-2 }[/tex]
= [tex]\frac{6-2\sqrt{2} )}{7 }[/tex]
Pelajari Lebih LanjutPenulis:
geniekvaa
Nilai jawaban:
0