Question

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

Answer 1

Answer:

bruh this is very easy

Step-by-step explanation:

given ,

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

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

Answer 2

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

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 :-

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

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

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

$\blue ➪ \: \: \sf \cfrac{34}{ - 23}$

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