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Mathematics

Find the square of:

5−x+2x5 - x +\dfrac{2}{x}

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Answer

(5−x+2x)2(5 - x +\dfrac{2}{x})^2

Using the formula

(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)

=(5)2+(x)2+(2x)2−2×5×x−2×x×(2x)+2×(2x)×5=25+x2+(4x2)−10x−(4xx)+(20x)=25+x2+(4x2)−10x−4+(20x)=(25−4)+x2+(4x2)−10x+(20x)=21+x2+(4x2)−10x+(20x)= (5)^2 + (x)^2 + \Big(\dfrac{2}{x}\Big)^2 - 2 \times 5 \times x - 2 \times x \times \Big(\dfrac{2}{x}\Big) + 2 \times \Big(\dfrac{2}{x}\Big) \times 5\\[1em] = 25 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x - \Big(\dfrac{4x}{x}\Big) + \Big(\dfrac{20}{x}\Big)\\[1em] = 25 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x - 4 + \Big(\dfrac{20}{x}\Big)\\[1em] = (25 - 4) + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x + \Big(\dfrac{20}{x}\Big)\\[1em] = 21 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x + \Big(\dfrac{20}{x}\Big)

Hence, (5−x+2x)2(5 - x +\dfrac{2}{x})^2 = 21 + x2 + (4x2)\Big(\dfrac{4}{x^2}\Big) - 10x + (20x)\Big(\dfrac{20}{x}\Big)

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