Find the square of:
2a−1a2a -\dfrac{1}{a}2a−a1
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(2a−1a)2(2a - \dfrac{1}{a})^2(2a−a1)2
Using the formula
(∵ (x - y)2 = x2 - 2xy + y2)
=(2a)2−2×2a×1a+1a2=4a2−4aa+1a2=4a2−4+1a2= (2a)^2 - 2 \times 2a \times \dfrac{1}{a} + \dfrac{1}{a}^2\\[1em] = 4a^2 - \dfrac{4a}{a} + \dfrac{1}{a^2}\\[1em] = 4a^2 - 4 + \dfrac{1}{a^2}=(2a)2−2×2a×a1+a12=4a2−a4a+a21=4a2−4+a21
Hence, (2a−1a)2(2a - \dfrac{1}{a})^2(2a−a1)2 = 4a2−4+1a24a^2 - 4 + \dfrac{1}{a^2}4a2−4+a21
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Expand:
(3x+13x)2\Big(3x +\dfrac{1}{3x}\Big)^2(3x+3x1)2
a+15aa +\dfrac{1}{5a}a+5a1
x - 2y + 1
3a - 2b - 5c
