Respuesta :

gmany

[tex]x^2+3x-10\neq0\\\\x^2+5x-2x-10neq0\\\\x(x-5)-2(x+5)\neq0\\\\(x+5)(x-2)\neq0\to x+5\neq0\ \wedge\ x-2\neq0\\\\x\neq-5\ \wedge\ x\neq2\\\\\dfrac{x+5}{x^2+3x-10}=\dfrac{x+5}{(x+5)(x-2)}=\dfrac{1}{x-2}\\\\Answer:\ x=2[/tex]

VirαtKσhli

Answer:

2

Step-by-step explanation:

Given function to us is ,

[tex]\implies f(x) =\dfrac{ x +5}{ x^2+3x -10}[/tex]

Here the denominator should not be equal to 0 . Therefore ,

[tex]\implies x^2 + 3x -10 \neq 0\\\\\implies ( x +5)(x-2) \neq 0 \\\\\implies x \neq 5 , -2 [/tex]

We can simplify the function and write it as ,

[tex]\implies f(x) =\dfrac{ x +5}{ x^2+3x -10}\\\\\implies \dfrac{ ( x +5)}{ ( x +5)(x-2)}\\\\\implies \dfrac{1}{x-2} [/tex]

Therefore our required answer is 2 .

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