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Natasha_Volkova [10]
2 years ago
11

Please help in the math

Mathematics
2 answers:
ICE Princess25 [194]2 years ago
8 0

Answer:

\rm \displaystyle \frac{x + y}{x - y} + \frac{x - y}{x + y} - \frac{2( {x}^{2} - {y}^{2}) }{ {x}^{2} - {y}^{2} } =  \boxed{ \displaystyle  \frac{4y ^2}{(x - y)(x + y)} }

Step-by-step explanation:

we want to simplify the following

\rm \displaystyle \frac{x + y}{x - y} + \frac{x - y}{x + y} - \frac{2( {x}^{2} - {y}^{2}) }{ {x}^{2} - {y}^{2} }

notice that we can reduce the fraction thus do so:

\rm \displaystyle \frac{x + y}{x - y} + \frac{x - y}{x + y} - \frac{2 \cancel{( {x}^{2} - {y}^{2}) }}{  \cancel{{x}^{2} - {y}^{2} }}

\rm \displaystyle \frac{x + y}{x - y} + \frac{x - y}{x + y} - 2

in order to simplify the addition of the algebraic fraction the first step is to figure out the LCM of the denominator and that is (x-y)(x+y) now divide the LCM by the denominator of very fraction and multiply the result by the numerator which yields:

\rm \displaystyle \frac{x + y}{x - y} + \frac{x - y}{x + y} - 2  \\   \\ \displaystyle  \frac{(x + y)^2 + (x - y)^2 - 2(x + y)(x - y)}{(x - y)(x + y)}

factor using (a-b)²=a²+b²-2ab

\rm \displaystyle  \frac{(x + y-(x - y) )^2}{(x - y)(x + y)}

remove parentheses

\rm \displaystyle  \frac{(x + y-x + y) )^2}{(x - y)(x + y)}

simplify:

\rm \displaystyle  \frac{4y ^2}{(x - y)(x + y)}

aivan3 [116]2 years ago
7 0

9514 1404 393

Answer:

  4y²/(x² -y²)

Step-by-step explanation:

The expression simplifies as follows:

  \dfrac{x+y}{x-y}+\dfrac{x-y}{x+y}-\dfrac{2(x^2-y^2)}{x^2-y^2}\\\\=\dfrac{(x+y)(x+y)+(x-y)(x-y)-2(x^2-y^2)}{(x-y)(x+y)}\\\\=\dfrac{(x+y)^2+(x-y)^2-2(x^2-y^2)}{x^2-y^2}\\\\=\dfrac{(x^2+2xy+y^2)+(x^2-2xy+y^2)-2(x^2-y^2)}{x^2-y^2}\\\\=\dfrac{2(x^2+y^2-(x^2-y^2))}{x^2-y^2}=\boxed{\dfrac{4y^2}{x^2-y^2}}

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Please help!! What Find the Error and explain why it is wrong!
Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply 2\cdot x -8 by the negative sign in numerator. Step E must be \frac{2\cdot x -5}{(x+4)\cdot (x-4)}.

Second image attached

The error was done in Step C, because the student omitted the 2\cdot a \cdot b of the algebraic identity (a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}. Step C must be 5\cdot x = x^{2}+4\cdot x + 4

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply 2\cdot x -8 by the negative sign in numerator. The real numerator in Step E should be:

3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x

Hence, Step E must be \frac{2\cdot x -5}{(x+4)\cdot (x-4)}.

Second image attached

The error was done in Step C, because the student omitted the 2\cdot a \cdot b of the algebraic identity (a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}. Step C must be 5\cdot x = x^{2}+4\cdot x + 4

And further steps are described below:

Step D

x^{2}-x+4 = 0

Which according to the Quadratic Formula, represents a polynomial with complex roots. That is: (a = 1, b = -1, c = 4)

D = b^{2}-4\cdot a\cdot c

D = (-1)-4\cdot (1)\cdot (4)

D = -17 (Conjugated complex roots)

Step E

(x-0.5-i\,1.936)\cdot (x-0.5+i\,1.936) = 0

Step F

x = 0.5+i\,1.936\,\lor\,x = 0.5-i\,1.936

8 0
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