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To solve the given problem, we need to find a common denominator for the fractions \(\frac{2}{2x+1}\) and \(\frac{1}{x+2}\). The common denominator for \(2x+1\) and \(x+2\) is \((2x+1)(x+2)\). Rewrite each fraction with the common denominator: \[ \frac{2}{2x+1} = \frac{2(x+2)}{(2x+1)(x+2)} \] \[ \frac{1}{x+2} = \frac{1(2x+1)}{(2x+1)(x+2)} \] Now, add the two fractions: \[ \frac{2(x+2)}{(2x+1)(x+2)} + \frac{2x+1}{(2x+1)(x+2)} = \frac{2(x+2) + (2x+1)}{(2x+1)(x+2)} \] Simplify the numerator: \[ 2(x+2) + (2x+1) = 2x + 4 + 2x + 1 = 4x + 5\] So, the expression becomes: \[ \frac{4x+5}{(2x+1)(x+2)} \] Therefore, the correct answer is: \[ \boxed{C} \]

Question

To solve the given problem, we need to find a common denominator for the fractions 22x+1\frac{2}{2x+1} and 1x+2\frac{1}{x+2}. The common denominator for 2x+12x+1 and x+2x+2 is (2x+1)(x+2)(2x+1)(x+2). Rewrite each fraction with the common denominator: 22x+1=2(x+2)(2x+1)(x+2) \frac{2}{2x+1} = \frac{2(x+2)}{(2x+1)(x+2)} 1x+2=1(2x+1)(2x+1)(x+2) \frac{1}{x+2} = \frac{1(2x+1)}{(2x+1)(x+2)} Now, add the two fractions: 2(x+2)(2x+1)(x+2)+2x+1(2x+1)(x+2)=2(x+2)+(2x+1)(2x+1)(x+2) \frac{2(x+2)}{(2x+1)(x+2)} + \frac{2x+1}{(2x+1)(x+2)} = \frac{2(x+2) + (2x+1)}{(2x+1)(x+2)} Simplify the numerator: 2(x+2)+(2x+1)=2x+4+2x+1=4x+5 2(x+2) + (2x+1) = 2x + 4 + 2x + 1 = 4x + 5 So, the expression becomes: 4x+5(2x+1)(x+2) \frac{4x+5}{(2x+1)(x+2)} Therefore, the correct answer is: C \boxed{C}

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Solution

Para resolver el problema dado, necesitamos encontrar un denominador común para las fracciones 22x+1\frac{2}{2x+1} y 1x+2\frac{1}{x+2}. El denominador común para 2x+12x+1 y x+2x+2 es (2x+1)(x+2)(2x+1)(x+2). Reescribimos cada fracción con el denominador común:

22x+1=2(x+2)(2x+1)(x+2) \frac{2}{2x+1} = \frac{2(x+2)}{(2x+1)(x+2)}

1x+2=1(2x+1)(2x+1)(x+2) \frac{1}{x+2} = \frac{1(2x+1)}{(2x+1)(x+2)}

Ahora, sumamos las dos fracciones:

2(x+2)(2x+1)(x+2)+2x+1(2x+1)(x+2)=2(x+2)+(2x+1)(2x+1)(x+2) \frac{2(x+2)}{(2x+1)(x+2)} + \frac{2x+1}{(2x+1)(x+2)} = \frac{2(x+2) + (2x+1)}{(2x+1)(x+2)}

Simplificamos el numerador:

2(x+2)+(2x+1)=2x+4+2x+1=4x+5 2(x+2) + (2x+1) = 2x + 4 + 2x + 1 = 4x + 5

Entonces, la expresión se convierte en:

4x+5(2x+1)(x+2) \frac{4x+5}{(2x+1)(x+2)}

Por lo tanto, la respuesta correcta es:

C \boxed{C}

This problem has been solved

Similar Questions

To solve the given problem, we need to find a common denominator for the fractions \(\frac{2}{2x+1}\) and \(\frac{1}{x+2}\). The common denominator for \(2x+1\) and \(x+2\) is \((2x+1)(x+2)\). Rewrite each fraction with the common denominator: \[ \frac{2}{2x+1} = \frac{2(x+2)}{(2x+1)(x+2)} \] \[ \frac{1}{x+2} = \frac{1(2x+1)}{(2x+1)(x+2)} \] Now, add the two fractions: \[ \frac{2(x+2)}{(2x+1)(x+2)} + \frac{2x+1}{(2x+1)(x+2)} = \frac{2(x+2) + (2x+1)}{(2x+1)(x+2)} \] Simplify the numerator: \[ 2(x+2) + (2x+1) = 2x + 4 + 2x + 1 = 4x + 5\] So, the expression becomes: \[ \frac{4x+5}{(2x+1)(x+2)} \] Therefore, the correct answer is: \[ \boxed{C} \]

\frac{3x+2}{4}=\frac{2x-1}{3}

limx--->1/2 (2x-1/4x^2-1)

Expressx(1 − 2x)2(1 − 3x)in partial fractions.

Dividing Fractions

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