Expand:
(3x+13x)2\Big(3x +\dfrac{1}{3x}\Big)^2(3x+3x1)2
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Using the formula
(∵ (x + y)2 = x2 + 2xy + y2)
=(3x)2+2×3x×13x+13x2=9x2+6x3x+19x2=9x2+2+19x2= (3x)^2 + 2 \times 3x \times \dfrac{1}{3x} + \dfrac{1}{3x}^2\\[1em] = 9x^2 + \dfrac{6x}{3x} + \dfrac{1}{9x^2}\\[1em] = 9x^2 + 2 + \dfrac{1}{9x^2}=(3x)2+2×3x×3x1+3x12=9x2+3x6x+9x21=9x2+2+9x21 Hence, (3x+13x)2=9x2+2+19x2\Big(3x +\dfrac{1}{3x}\Big)^2 = 9x^2 + 2 + \dfrac{1}{9x^2}(3x+3x1)2=9x2+2+9x21
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(a + b - c)2
(a - b + c)2
Find the square of:
a+15aa +\dfrac{1}{5a}a+5a1
2a−1a2a -\dfrac{1}{a}2a−a1
