The value of x-4 2x 2x2x x-4 2x2x 2x x-4- x -45-4 is

Applying C_1→ C_1+C_2+C_3, We get, Δ= 5x+4 amp;2x amp;2x 5x+4 amp;x+4 amp;2x 5x+4 amp;2x amp;x+4 =(5x+4) 1 amp;2x amp;2x 1 amp;x+4 a ...

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Modulus of a Complex Number

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The value of $∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣, x \in R - {\frac{- 4}{5}, 4}$ is

(5 x + 4) (4 - x)

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(5 x - 4) (4 - x)^{2}

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(5 x + 4) (4 - x)^{2}

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(5 x + 4) (4 - x)^{3}

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$∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣ = (5 x + 4) (4 - x)^{2}$ .

$∣ ∣ ∣ ∣ \begin{matrix} y + k & y & y y & y + k & y y & y & y + k \end{matrix} ∣ ∣ ∣ ∣ = k^{2} (3 y + k)$

x

(x^{2} - 2 x) (2 x - 2) - 9 \frac{2 x - 2}{x^{2} - 2 x} \leq 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{2} x}{3 {cos}^{2} x}, a, \frac{b (1 - sin x)}{{(π - 2 x)}^{2}} \end{matrix} \begin{matrix} x < \frac{π}{2} x = \frac{π}{2} x > \frac{π}{2} \end{matrix}

f (x)

x = \frac{π}{2}

2 x (2 x - 3) + 5

x = 2

∣ ∣ ∣ ∣ \begin{matrix} x^{2} + 2 x & 2 x + 1 & 1 2 x + 1 & x + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = {(x - 1)}^{3} .

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