(a−12a)2\Big(a - \dfrac{1}{2a}\Big)^2(a−2a1)2 is equal to :
(a2+14a2−2)\Big(a^2 + \dfrac{1}{4a^2} - 2\Big)(a2+4a21−2)
(a2+14a2+2)\Big(a^2 + \dfrac{1}{4a^2} + 2\Big)(a2+4a21+2)
(a2+14a2−1)\Big(a^2 + \dfrac{1}{4a^2} - 1\Big)(a2+4a21−1)
(a2−14a2−2)\Big(a^2 - \dfrac{1}{4a^2} - 2\Big)(a2−4a21−2)
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Given,
⇒ (a−12a)2\Big(a - \dfrac{1}{2a}\Big)^2(a−2a1)2
Expanding,
⇒(a−12a)(a−12a)⇒a2−a×12a−12a×a−12a×−12a⇒a2−12−12+14a2⇒a2+14a2−1.\Rightarrow \Big(a - \dfrac{1}{2a}\Big)\Big(a - \dfrac{1}{2a}\Big) \\[1em] \Rightarrow a^2 - a \times \dfrac{1}{2a} - \dfrac{1}{2a} \times a - \dfrac{1}{2a} \times -\dfrac{1}{2a} \\[1em] \Rightarrow a^2 - \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{4a^2} \\[1em] \Rightarrow a^2 + \dfrac{1}{4a^2} - 1.⇒(a−2a1)(a−2a1)⇒a2−a×2a1−2a1×a−2a1×−2a1⇒a2−21−21+4a21⇒a2+4a21−1.
Hence, Option 3 is the correct option.
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x2 - (a + b)x + ab is the expansion of :
(x - b)(x - a)
(x - b)(x + a)
(x + b)(x - a)
(x + a)(x + b)
If a + b - c = 4 and a2 + b2 + c2 = 14, the value of ab - bc - ca is :
2
1
0.5
-0.5
Expand (x + 8)(x + 10)
Expand (x + 8)(x - 10)
