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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]


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