Question

Expand (2x-1/3x) ^2​

Answer 1

Step by step Expansion :

${( \frac{2x - 1}{3x} })^{2} \\ \\ = \frac{ {(2x)}^{2} + {1}^{2} - 2x}{9x} \\ \\ = \frac{4 {x}^{2} - 2x + 1 }{9x}$

Answer 2

Answer:

$4x - (\frac{4}{3}) + \frac{1}{9x^{2} }$

Given:

$(2x - \frac{1}{3x} )^{2}$

Step-by-step Explanation:

According to identity:

$(a - b)^{2} = a^{2} - 2ab + b^{2}$

using this identity, we can write

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

⇒ $(2x - \frac{1}{3x} )^{2} = 4x - (\frac{4}{3}) + \frac{1}{9x^{2} }$

Hence, $(2x - \frac{1}{3x} )^{2} = 4x - (\frac{4}{3}) + \frac{1}{9x^{2} }$.

#SPJ2