Answered

IDNStudy.com, kung saan ang iyong mga tanong ay natutugunan ng mga eksperto. Alamin ang mga maaasahang sagot sa iyong mga tanong mula sa aming malawak na kaalaman sa mga eksperto.

Kindly solve the following:
(16x^-6 y^4)^1/2

(8x^-3 y^-9)^2/3

(-2a^3/2 b^-1)^4

(a^1/2 - b^1/2)^2

Thanks!

Sagot :

Kaiusidn
1. First, we expand the expression: ((4^2)(y^4) / (x^6))^1/2 --> By laws of exponents, we multiply 1/2 by the exponents of the terms: [(4^2/2)(y^4/2) / (x^6/2)] = (4y^2 / x^3) 2. We first expand the expression: [(2^3) / (x^3)(y^9)]^2/3 --> By laws of exponents, we multiply thte exponenets of the terms by 2/3: [(2^6/3) / (x^6/3)(y^18/3)] = (4 / x^2 y^6) 3. Since there are no fraction exponents, we now proceed into multiplying the exponents: [((-2)^4)(a^12/2) / (b^4)] = (16a^6 / b^4) 4. Since there is an operation, we first expand the term: (a^1/2 - b^1/2)(a^1/2 - b^1/2) = (a^2/2 - 2(a^1/2)(b^1/2) + b^2/2) = a - 2(a^1/2)(b^1/2) + b
Riza1idn
[tex]1)\\\\(16x^{-6} y^4)^{\frac{1}{2}}=16^{\frac{1}{2}} \cdot x^{-6\cdot \frac{1}{2}}\cdot y^{4\cdot\frac{ 1}{2}}=\sqrt{16}\cdot x^{-3}y^{2}=4x^{-3}y^{2}=\frac{1}{x^3}\cdot 4y^{2}=\frac{4y^2}{x^3}\\\\x\neq 0[/tex]


[tex]2)\\\\(8x ^{-3} y^{-9} )^{\frac{2}{3}} =8^{\frac{2}{3}} \cdot x^{-3\cdot \frac{2}{3}}\cdot y^{-9\cdot\frac{ 2}{3}}=\sqrt[3]{8^{2}} \cdot x^{-2} \cdot y^{-6} =\sqrt[3]{64} \cdot\frac{1}{x^{ 2}} \cdot \frac{1}{y^{ 6}} =\frac{4}{x^2y^6}\\\\x,y\neq0[/tex]
 

[tex]3)\\\\(-2a^{\frac{3}{2}} \cdot b^{-1 })^4=(-2)^{4}\cdot a^{\frac{3}{2}\cdot 4} \cdot b^{-1\cdot 4}=(-2)^{4}\cdot a^{\frac{3}{2}\cdot 4} \cdot b^{(-1)\cdot 4}=16a^{6} \cdot b^{-4}=\\\\=16a^{6} \cdot \frac{1}{b^{ 4}}=\frac{16a^6}{b^{4}}\\\\b\neq 0[/tex]


[tex]4)\\\\(a^{\frac{1}{2}} - b ^{\frac{1}{2}})^2= (\sqrt{a}-\sqrt{b})^{2}=(\sqrt{a})^{2}-2\sqrt{a}\sqrt{b}+(\sqrt{b})^{2}=a-2\sqrt{ab}+b[/tex]