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Given \(3x+\dfrac{2}{x}-4=2(x+2)-\left(9-\dfrac{1}{x}\right)\) what is \(\dfrac{x^2+1}{x}?\)

 Jun 24, 2022

Best Answer 

 #1
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Note that \({{x^2 + 1} \over x} = {x^2 \over x} + {1 \over x} = x + {1 \over x}\)

 

Now, simplify the equation to: \(3x + {2 \over x} - 4 = 2x -5+ {1 \over x}\)

 

Now, subtract \(2x \) from both sides: \(x + {2 \over x} - 4 = -5 + {1 \over x}\)

 

Subtract \({1 \over x}\) from both sides: \(x + {1 \over x} - 4 = -5\)

 

Can you take it from here?

 Jun 24, 2022
 #1
avatar+2448 
0
Best Answer

Note that \({{x^2 + 1} \over x} = {x^2 \over x} + {1 \over x} = x + {1 \over x}\)

 

Now, simplify the equation to: \(3x + {2 \over x} - 4 = 2x -5+ {1 \over x}\)

 

Now, subtract \(2x \) from both sides: \(x + {2 \over x} - 4 = -5 + {1 \over x}\)

 

Subtract \({1 \over x}\) from both sides: \(x + {1 \over x} - 4 = -5\)

 

Can you take it from here?

BuilderBoi Jun 24, 2022

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