-x-3-x-x-2-1- then x -B- 1-5C- 1-5D- -5-1

The correct option is D { 5, 1}|x+3|+xx+2=1, x≠ 2 ⇒|x+3| 2x+2=0 For x ≥ 3, x+1x+2=0 ⇒ x= 1 For x 3, x+5x+2=0 ⇒ x= 5 ∴ x= 5, 1 nbsp; ...

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Absolute Value Function

|x+3|+x/x+2=1...

Question

If $\frac{| x + 3 | + x}{x + 2} = 1$ , then $x \in$

{- 5, - 1}

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{1, 5}

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{- 1, 5}

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{1, 5}

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\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)}

+ \frac{C}{(x + 1) (x + 2) (x + 3)},

B

f (x)

x = - 2

x = - 1

x = \frac{1}{3}

\int x^{5} (1 + x^{3})^{2 / 3} d x = A (1 + x^{3})^{8 / 3} + B (1 + x^{3})^{5 / 3} + c

x^{3} - \frac{1}{x^{3}} = 14

x - \frac{1}{x} =

\frac{x^{2} + 3 x + 5}{(x + 1) (x + 2) (x + 3)} =

\frac{A}{(x + 1)} + \frac{B}{(x + 1) (x + 2)} + \frac{c}{(x + 1) (x + 2) (x + 3)}

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