Jadikan ke bentuk paling sederhana
[tex]\large \frac{\left ( x^{\frac{1}{3}}-x^{\frac{1}{6}} \right )\left ( x^{\frac{1}{2}}+x \right )\left ( x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}{\left ( x^{\frac{4}{3}}-x \right )\left ( x+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}[/tex]
[tex]\large \frac{\left ( x^{\frac{1}{3}}-x^{\frac{1}{6}} \right )\left ( x^{\frac{1}{2}}+x \right )\left ( x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}{\left ( x^{\frac{4}{3}}-x \right )\left ( x+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}[/tex]
Verified answer
Bentuk paling sederhana dari [tex]\tt \frac{(x^\frac{1}{3}-x^\frac{1}{6})(x^\frac{1}{2}+x)(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{(x^\frac{4}{3}-x)(x+x^\frac{1}{3}+x^\frac{2}{3})}[/tex] adalah [tex]\tt \frac{\sqrt[3]{x^2}}{x}[/tex]/disederhanakan lagi menjadi [tex]\tt x^{-\frac{1}{3}}[/tex].
Pembahasan
Eksponen merupakan metode yang digunakan untuk mengganti pengulangan maupun perpangkatan angka. Angka pada bentuk eksponensial disebut basis, dan angka yang menyatakan berapa kali basis berulang disebut eksponen.
Sifat eksponen :
[tex]\tt 1. (\frac{a^m}{b^n})^p= \frac{a^{mp}}{b^{np}}\\ \\ 2. a^0=1\\\\3.(a^mb^n)^p=a^{mp}.b^{np}\\\\4. a^\frac{m}{n}= \sqrt[n]{a^m}\\ \\ 5. a^n= \frac{1}{a^{-n}} \\\\6. a^m\times a^n=a^{m+n}\\\\7. a^m\div a^n=a^{m-n}\\\\8. (a^m)^n =a^{mn}[/tex]
Akar adalah metode kebalikan pada eksponen.
Sifat bentuk akar :
[tex]\tt 1. \sqrt{a} \times \sqrt{a} =(\sqrt{a})^2=a\\\\2. \frac{\sqrt[n]{a}}{\sqrt[n]{b}} =\sqrt[n]{\frac{a}{b}} \\\\3. \sqrt[n]{a^{mn+p}.b^q}=a^m.\sqrt[n]{a^p.b^q} \\\\4. x\sqrt[n]{a} \times y\sqrt[n]{b}= xy\sqrt[n]{ab}\\ \\ 5. (\sqrt{a} +\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b\\ \\ 6.(\sqrt{a} \pm\sqrt{b})^2=(a+b)\pm 2\sqrt{ab}\\ \\7.\sqrt[n]{a} \times \sqrt[n]{b} =\sqrt[n]{ab}[/tex]
Penyelesaian Soal
Diketahui :
[tex]\tt \frac{(x^\frac{1}{3}-x^\frac{1}{6})(x^\frac{1}{2}+x)(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{(x^\frac{4}{3}-x)(x+x^\frac{1}{3}+x^\frac{2}{3})}[/tex]
Ditanya :
Bentuk paling sederhana...?
Jawaban :
[tex]\tt \frac{(x^\frac{1}{3}-x^\frac{1}{6})(x^\frac{1}{2}+x)(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{(x^\frac{4}{3}-x)(x+x^\frac{1}{3}+x^\frac{2}{3})}\\\\= \frac{x^\frac{1}{6}(x^\frac{1}{6}-1)x^\frac{1}{2}(1+x^\frac{1}{2})(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{(\sqrt[3]{x^4}-x)x^\frac{1}{6}(x^\frac{5}{6}+x^\frac{1}{6} +x^\frac{1}{2})}\\ \\[/tex]
[tex]\tt = \frac{(x^\frac{1}{6} -1)x^\frac{1}{2}(1+x^\frac{1}{2})(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{(\sqrt[3]{x^3} \sqrt[3]{x})x^\frac{1}{6}(x^\frac{2}{3}+1+x^\frac{1}{3})} \\\\= \frac{(x^\frac{1}{6}-1)x^{\frac{1}{2}-\frac{1}{6}}(1+x^\frac{1}{2})(x^\frac{1}{2}+x^\frac{1}{3} +x^\frac{2}{3})}{x(\sqrt[3]{x} -1)(x^\frac{2}{3} +1+x^\frac{1}{3})}\\ \\[/tex]
[tex]\tt = \frac{(x^\frac{1}{2} +x^\frac{1}{2}\times x^\frac{1}{2}-x^\frac{1}{3}-x^\frac{1}{3}\times x^\frac{1}{2})(x^\frac{1}{2}+x^\frac{1}{3}+x^\frac{2}{3})}{x(\sqrt[3]{x}-1)(\sqrt[3]{x^2} +1+\sqrt[3]{x}} \\\\=\frac{x^\frac{5}{6} +x^\frac{7}{6} +x^\frac{3}{2} +x^\frac{4}{3} +x^\frac{5}{3}-x^\frac{5}{6} -x^\frac{2}{3} -x^\frac{4}{3}-x^\frac{7}{6}-x^\frac{3}{2}}{x(\sqrt[3]{x} -1)(\sqrt[3]{x^2}+1+\sqrt[3]{x})} \\\\[/tex]
[tex]\tt = \frac{\sqrt[3]{x^3} \sqrt[3]{x^2}-\sqrt[3]{x^2}}{x(\sqrt[3]{x^2}-1)(\sqrt[3]{x^2}+1+\sqrt[3]{x})}\\ \\= \frac{\sqrt[3]{x^2}(x-1)}{x(\sqrt[3]{x}-1)(\sqrt[3]{x^2} +1+\sqrt[3]{x})}\\ \\= \frac{\sqrt[3]{x^2}(\sqrt[3]{x} -1)((\sqrt[3]{x})^2+\sqrt[3]{x}+1^2)}{x(\sqrt[3]{x}-1)(\sqrt[3]{x^2} +1+\sqrt[3]{x})} \\\\= \frac{\sqrt[3]{x^2}}{x}\\\\=x^{\frac{2}{3} -1}\\\\= x^\frac{2-3}{3}\\ \\=x^{-\frac{1}{3}}[/tex]
Pelajari Lebih Lanjut
Detail Jawaban
Kelas : IX
Mapel : Matematika
Kategori : 1 - Perpangkatan dan bentuk akar
Kode : 9.2.1
Kata kunci : Perpangkatan dan Bilangan berpangkat, Bentuk akar, Bentuk paling sederhana