Expand (2x−1x)(3x+2x)\Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big)(2x−x1)(3x+x2)
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Given,
(2x−1x)(3x+2x)\Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big)(2x−x1)(3x+x2)
Expanding,
⇒2x×3x+2x×2x−1x×3x−1x×2x⇒6x2+4−3−2x2⇒6x2+1−2x2.\Rightarrow 2x \times 3x + 2x \times \dfrac{2}{x} - \dfrac{1}{x} \times 3x - \dfrac{1}{x} \times \dfrac{2}{x} \\[1em] \Rightarrow 6x^2 + 4 - 3 - \dfrac{2}{x^2} \\[1em] \Rightarrow 6x^2 + 1 - \dfrac{2}{x^2}.⇒2x×3x+2x×x2−x1×3x−x1×x2⇒6x2+4−3−x22⇒6x2+1−x22.
Hence, (2x−1x)(3x+2x)=6x2+1−2x2.\Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big) = 6x^2 + 1 - \dfrac{2}{x^2}.(2x−x1)(3x+x2)=6x2+1−x22.
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