Penjelasan dengan langkah-langkah: Oke, jika diketahui [tex]\displaystyle x=\frac{1}{4},\:\:y=2,\:\:z = 1[/tex] Maka nilai dari [tex]\displaystyle\frac{x^{-2}yz^2}{xy^2z^{-1}}\\\\\because n^{-a}=\frac{1}{n^a},\:\:\:\frac{1}{n^{-a}}=n^a\therefore\\\\\frac{x^{-2}yz^2}{xy^2z^{-1}}=\frac{yz^2z^{1}}{x^{2}xy^2}\\\\\because n^an^b=n^{a+b}\therefore\\\\\frac{yz^2z^{1}}{x^{2}xy^2}=\frac{yz^{2+1}}{x^{2+1}y^2}=\frac{yz^{3}}{x^{3}y^2}\\\\\because\frac{n^a}{n^b}=n^{a-b},\:\:\:\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\therefore\\\\\frac{yz^{3}}{x^{3}y^2}=\frac{y}{y^2}\left(\frac{z^{3}}{x^{3}}\right)[/tex]
Jawab:
= 32
Penjelasan dengan langkah-langkah:
Oke, jika diketahui
[tex]\displaystyle x=\frac{1}{4},\:\:y=2,\:\:z = 1[/tex]
Maka nilai dari
[tex]\displaystyle\frac{x^{-2}yz^2}{xy^2z^{-1}}\\\\\because n^{-a}=\frac{1}{n^a},\:\:\:\frac{1}{n^{-a}}=n^a\therefore\\\\\frac{x^{-2}yz^2}{xy^2z^{-1}}=\frac{yz^2z^{1}}{x^{2}xy^2}\\\\\because n^an^b=n^{a+b}\therefore\\\\\frac{yz^2z^{1}}{x^{2}xy^2}=\frac{yz^{2+1}}{x^{2+1}y^2}=\frac{yz^{3}}{x^{3}y^2}\\\\\because\frac{n^a}{n^b}=n^{a-b},\:\:\:\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\therefore\\\\\frac{yz^{3}}{x^{3}y^2}=\frac{y}{y^2}\left(\frac{z^{3}}{x^{3}}\right)[/tex]
[tex]\displaystyle=y^{1-2}\left(\frac{z}{x}\right)^3=y^{-1}\left(\frac{z}{x}\right)^3\\\\=\frac{1}{y}\left(\frac{z}{x}\right)^3[/tex]
Masukkan nilai x, y dan z
sesuai dengan yang diketahui
[tex]\displaystyle \frac{1}{y}\left(\frac{z}{x}\right)^3\\\\=\frac{1}{2}\left(\frac{1}{\frac{1}{4}}\right)^3\\\\\\\because\:\frac{1}{\frac{1}{a}}=a\:\:\{a\ne0\}\:\therefore\\\\\\=\frac{1}{2}\left(4\right)^3\\\\=\frac{1}{2}\left(64\right)[/tex]
= 32
(xcvi)