Let α and β be the roots of the given quadratic equation
x² + 8x - 65 = 0
a = 1, b = 8, c = -65
α + β = \(-\frac {b}{a}\) = -8 and αβ = \(\frac {c}{a}\) = -65
α + β = -8; αβ = -65

General form of the quadratic equation when the roots are given is
x² - (sum of the roots)x + product of the roots = 0
x² - 9x + 14 = 0

Let p(x) = x² + 18x + 12 = (x + 2)(x + 6)
p(-2) = 4 - 16 + 20 = 0
p(-6) = 36 - 48 + 12 = 0
Therefore -2 and -6 are zero of p(x) = x² + 8x + 12

|2x + 5|

\(\sqrt { 256{ \left( x-a \right) }^{ 8 }{ \left( x-b \right) }^{ 4 }{ \left( x-c \right) }^{ 16 }{ \left( x-d \right) }^{ 20 } } \) = 16|(x - a)⁴(x - b)² (x - c)⁸ (x - d)¹⁰|

\(\frac { x\left( x+1 \right) }{ x-2 } +\frac { x\left( 1-x \right) }{ x-2 } =\frac { x(x+1)+x(x(1-x) }{ (x-2) } =\frac { 2x }{ x-2 } \)

\(\frac { 1 }{ { x }^{ 2 }-5x+6 } +\frac { 1 }{ { x }^{ 2 }-3x+2 } -\frac { 1 }{ { x }^{ 2 }-8x+15 } \)
= \(\frac { 1 }{ \left( x-2 \right) \left( x-3 \right) } +\frac { 1 }{ \left( x-2 \right) \left( x-1 \right) } -\frac { 1 }{ \left( x-5 \right) \left( x-3 \right) } \)
= \(\frac { \left( x-1 \right) \left( x-5 \right) +\left( x-3 \right) \left( x-5 \right) -\left( x-1 \right) \left( x-2 \right) }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) \left( x-5 \right) } \)
= \(\frac { { x }^{ 2 }-11x+18 }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) \left( x-5 \right) } =\frac { \left( x-9 \right) \left( x-2 \right) }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) \left( x-5 \right) } \)
= \(\frac { x-9 }{ \left( x-1 \right) \left( x- \right) \left( x-5 \right) } \)

\(\frac { 14{ x }^{ 4 } }{ y } \div \frac { 7x }{ 3{ y }^{ 4 } } =\frac { 14{ x }^{ 4 } }{ y } \times \frac { 3{ y }^{ 4 } }{ 7x } \) = 6x³y³

\(\frac { { x }^{ 3 } }{ 9{ y }^{ 2 } } \times \frac { 27y }{ { x }^{ 5 } } =\frac { 3 }{ { x }^{ 2 }y } \)

\(\frac { y }{ { y }^{ 2 }-25 } =\frac { y }{ (y+5)(y-5) } \) is undefined when (y+5)(y-5)=0 that is y=-5,5.
∴ The excluded values are -5,5.

\(\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+x } =\frac { (x+1)(x-1) }{ { x }(x+1) } =\frac { x-1 }{ x } \)

Let f(x) = 21x²y
g(x) = 35 xy²
GCD = 7 xy
LCM of 21, 35 = 105
LCM of x²y, xy² = x²y²
LCM = 105 x²y²
Now, f(x) x g(x) = (21 x²y) (35 xy²)
= 735 x³ y³
LCM x GCD = (105 x²y²) (7 xy)
= 735 x³ y³
f(x) x g(x) = LCM x GCD
Hence verified

4x²y, 8x³y²
LCM of (4, 8) = 8
LCM of (x²y, x³y²) = x³y²
LCM of (4x²y, 8x³y²) = 8x³y²

8x⁴y², 48x²y⁴
First let us find the LCM of the numerical coefficients.
That is, LCM (8, 48) = 2 x 2 x 2 x 6 = 48
Then find the LCM of the terms involving variables.
That is, LCM (x⁴y², x²y⁴) = x⁴y⁴
Finally find the LCM of the given expression.
We conclude that the LCM of the given expression is the product of the LCM of the numerical coefficient and the LCM of the terms with variables.
Therefore, LCM (8x⁴y², 48x²y⁴) = 48 x⁴y⁴

2x − 3y = 6 … (1)
x + y = 1 … (2)


Substituting, y = \(\frac {-4}{5}\) in (2), x - \(\frac {4}{5}\) = 1 we get, x = \(​​\frac {9}{5}\)
Therefore, x = \(​​\frac {9}{5}\), y = \(\frac {-4}{5}\).