Math, asked by pjahnabi007, 3 months ago

a=1, b=3, and c= -2, find the value of​

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Answers

Answered by Failboat
2

Answer:

bruh this is very easy

Step-by-step explanation:

given ,

a=1, b=3, and c=-2

\frac{2a^2+4b^2-c^2}{a^2 b - 2ab^2 - 2ac^2} = \frac{2(1)^2+4(2)^2 - (-2)^2}{(1)^2 (3) - 2(1)(3)^2 - 2(1)(-2)^2} = \frac{2+16-4}{3-18-8} = - \frac{14} {23}

Answered by srikanthn711
17

\large \mathfrak \red { ☄\:  \: Given \:  \: ☄}

The values of :-

  • a = 1
  • b = 3
  • c = -2

 \\

\large \mathfrak \red {☄ \:  \: To \: find \:  \:☄ }

We have to find the value of :-

\sf \dfrac{2 {a}^{2}  + 4 {b}^{2}  -  {c}^{2} }{ {a}^{2}b  - 2a {b}^{2} - 2a {c}^{2}  }

 \\

\large \mathfrak \red {☄ \:  \: Solution \:  \: ☄}

Substituting the values and solving :-

\blue ➪ \:  \: \sf    \cfrac{2(1)^2+4(3)^2-(-2)^2}{(1)^2(3)-2(1)(3)^2-2(1)(-2)^2}

\blue ➪ \:   \:  \sf  \cfrac{2(1) + 4(9) - (4)}{1(3) - 2(9) - 2(4)}

\blue ➪ \:  \: \sf  \cfrac{2+36 - 4}{3 - 18 - 8}

\blue ➪ \:  \: \sf  \cfrac{34}{ - 23}

\sf \blue {\therefore  \:  \dfrac{2a^2+4b^2-c^2}{a^2b-2ab^2-2ac^2}= \dfrac{34}{-23}  }

 \\

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