The solution to each of the question takes different approach, as the questions are taken from different concepts; however, a common operation among all questions, is factorization.
[tex](1)\ (-xy)^3(xz)[/tex]
Expand
[tex](-xy)^3(xz) = (-x)^3* y^3*(xz)[/tex]
[tex](-xy)^3(xz) = -x^3* y^3*xz[/tex]
Rewrite as:
[tex](-xy)^3(xz) = -x^3*x* y^3*z[/tex]
Apply law of indices
[tex](-xy)^3(xz) = -x^4y^3z[/tex]
[tex](2)\ (\frac{1}{3}mn^{-4})^2[/tex]
Expand
[tex](\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}[/tex]
[tex](\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8[/tex]
[tex](3)\ (\frac{1}{5x^4})^{-2}[/tex]
Apply negative power rule of indices
[tex](\frac{1}{5x^4})^{-2}= (5x^4)^2[/tex]
Expand
[tex](\frac{1}{5x^4})^{-2}= 5^2x^{4*2}[/tex]
[tex](\frac{1}{5x^4})^{-2}= 25x^{8[/tex]
[tex](4)\ -x(2x^2 - 4x) - 6x^2[/tex]
Expand
[tex]-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2[/tex]
Evaluate like terms
[tex]-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2[/tex]
Factor out x^2
[tex]-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2[/tex]
Factor out -2
[tex]-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2[/tex]
[tex](5)\ \sqrt{\frac{4y}{3y^2}}[/tex]
Divide by y
[tex]\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}[/tex]
Split
[tex]\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}[/tex]
[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}[/tex]
Rationalize
[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}[/tex]
[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}[/tex]
[tex](6)\ \frac{8}{3 + \sqrt 3}[/tex]
Rationalize
[tex]\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}[/tex]
[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}[/tex]
Apply difference of two squares to the denominator
[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}[/tex]
[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}[/tex]
[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}[/tex]
Simplify
[tex]\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}[/tex]
[tex](7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}[/tex]
Expand
[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}[/tex]
Split
[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}[/tex]
Evaluate all roots
[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}[/tex]
[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}[/tex]
[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}[/tex]
[tex](8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}[/tex]
Expand
[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}[/tex]
Factorize each
[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}[/tex]
Factor out (r+3) in the numerator and (r + 6) in the denominator
[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}[/tex]
Cancel out r - 2
[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}[/tex]
[tex](9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}[/tex]
Cancel out x
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}[/tex]
Expand the numerator of the 2nd fraction
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}[/tex]
Factorize
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}[/tex]
Factor out x - 7
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}[/tex]
Factor out 4 from 4x + 8
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}[/tex]
Cancel out x + 2
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}[/tex]
[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}[/tex]
[tex](10)\ (3x^3 + 15x^2 -21x) \div 3x[/tex]
Factorize
[tex](3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x[/tex]
Cancel out 3x
[tex](3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7[/tex]
[tex](11)\ \frac{m}{6m + 6} - \frac{1}{m+1}[/tex]
Take LCM
[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}[/tex]
Expand
[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}[/tex]
[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}[/tex]
[tex](12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}[/tex]
Rewrite as:
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}[/tex]
Express as multiplication
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}[/tex]
Express y^2 - 9 as y^2 - 3^2
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}[/tex]
Express as difference of two squares
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}[/tex]
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}[/tex]
[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}[/tex]
Read more at:
https://brainly.com/question/4372544
