For all real numbers x such that x \neq 0, the expression \left(\frac{1}{x} + 1\right)^2 -… — ACT

For all real numbers $x$ such that $x \neq 0$ , the expression $\left(\frac{1}{x} + 1\right)^2 - \frac{1}{x^2}$ is equivalent to which of the following?

$\frac{x+2}{x}$ Cevap
B
$1$
C
$\frac{3}{x^2}$
D
$\frac{x+2}{2x}$
E
$2x$

Cevap

\frac{x+2}{x}

Expanding the squared term yields

\frac{1}{x^2} + \frac{2}{x} + 1

. Subtracting

\frac{1}{x^2}

from this leaves

\frac{2}{x} + 1

. Finding a common denominator of

x

to add these terms results in

\frac{x+2}{x}

Adım Adım Çözüm

Expand the binomial term

\left(\frac{1}{x} + 1\right)^2

\frac{1}{x^2} + \frac{2}{x} + 1

Using the algebraic identity

(a+b)^2 = a^2 + 2ab + b^2

where

a = \frac{1}{x}

and

b = 1

Substitute this expansion back into the original expression and subtract

\frac{1}{x^2}

\frac{2}{x} + 1

The positive and negative

\frac{1}{x^2}

terms cancel each other out.

Write the expression as a single fraction over the common denominator

x

\frac{x+2}{x}

Convert

1

\frac{x}{x}

and add the numerators.

Anahtar Kavram

Simplifying rational expressions by expanding binomials and finding common denominators.

Alternatif Yöntem

Use the difference of squares factorization:

A^2 - B^2 = (A-B)(A+B)

. Let

A = \frac{1}{x} + 1

and

B = \frac{1}{x}

. Then the expression simplifies to

\left(\frac{1}{x} + 1 - \frac{1}{x}\right)\left(\frac{1}{x} + 1 + \frac{1}{x}\right) = 1 \cdot \left(\frac{2}{x} + 1\right) = \frac{x+2}{x}

Tahmini Süre:1m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Konular

Bu konudan daha fazla soru