1.tentukanlah nilai dari a. (x-y)^-3 2.tentukan nilai positif dan sederhanakan dari a. {3a^2b^-3/a^-5b}^-2 b. [tex] \frac{x^{-2} + y^{-2} }{x^{-1} + y^{-1

[tex]\displaystyle 1\\(x-y)^{-3}=\frac1{(x-y)^3}\\\boxed{\boxed{(x-y)^{-3}=\frac1{x^3-3x^2y+3xy^2-y^3}}}\\\\\\2\\\left(\frac{3a^2b^{-3}}{a^{-5}b}\right)^{-2}=\left(3a^{2-(-5)}b^{-3-1}}\right)^{-2}\\\left(\frac{3a^2b^{-3}}{a^{-5}b}\right)^{-2}=\left(3a^{7}b^{-4}}\right)^{-2}\\\left(\frac{3a^2b^{-3}}{a^{-5}b}\right)^{-2}=3^{-2}a^{7\cdot(-2)}b^{-4\cdot(-2)}\\\left(\frac{3a^2b^{-3}}{a^{-5}b}\right)^{-2}=3^{-2}a^{-14}b^{8}\\\boxed{\boxed{\left(\frac{3a^2b^{-3}}{a^{-5}b}\right)^{-2}=\frac{b^{8}}{9a^{14}}}}[/tex]

[tex]\displaystyle \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}=\frac{\frac1{x^2}+\frac1{y^2}}{\frac1{x}+\frac1{y}}\\\frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}=\frac{\frac{y^2+x^2}{x^2y^2}}{\frac{y+x}{xy}}\\\frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}=\frac{\frac{y^2+x^2}{xy}}{y+x}\\\boxed{\boxed{\frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}=\frac{y^2+x^2}{y^2x+x^2y}}}[/tex]

Nomor 1
= (x - y)`³
= 1 / (x - y)³
= 1 / (x - y) (x - y) (x - y)
= 1 / (x³ - 3x²y + 3xy² - y³)

Nomor 2
Bagian A
= (3a² b`³/ a^-5 b)`²
= (3a^(2+5) b^(-3-1) )`²
= (3a^7 b`⁴)`²
= 3`² a^-14 b^8
= b^8 / (9a^14)

Bagian B
= (x`² + y`²) / (x`¹ + y`¹)
= (1/x² + 1/y²) : (1/x + 1/y)
= (y² + x²) / x²y² : (y+x) / xy
= (x² + y²) / x²y² × xy / (x + y)
= xy (x² + y²) / x²y²(x + y)
= (x² + y²) / (x²y + y²x)

semoga berguna +_+