Darurat!!! Tolonggg dibantu pakai caraa
Matematika
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Pertanyaan
Darurat!!! Tolonggg dibantu pakai caraa
1 Jawaban
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1. Jawaban Yoono
[tex] \sqrt[5]{2x^{3}.x^{1/2}} . x^{1/2}\sqrt{3} \\ \sqrt[5]{2x^{7/2}} . x^{1/2}\sqrt{3} \\ \sqrt[5]{2}.x^{7/10} . x^{1/2}\sqrt{3} \\ x^{7/20}\sqrt[5]{2}\sqrt{3}[/tex]