x2+1x2−3x^2 + \dfrac{1}{x^2} - 3x2+x21−3 in the form of factors is :
(x−1x)(x+1x)\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)(x−x1)(x+x1) - 1
(x+1x+1)(x−1x−1)\Big(x + \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x+x1+1)(x−x1−1)
(x−1x)(x−1x−1)\Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1)(x−x1−1)
(x−1x+1)(x−1x−1)\Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1+1)(x−x1−1)
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Given,
=x2+1x2−3=x2+1x2−2−1=x2+1x2−(2×x×1x)−1=[x2+1x2−(2×x×1x)]−1=(x−1x)2−1=(x−1x)2−12=(x−1x+1)(x−1x−1).\phantom{=}x^2 + \dfrac{1}{x^2} - 3 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 1 \\[1em] = x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big) - 1 \\[1em] = \Big[x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big)\Big] - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).=x2+x21−3=x2+x21−2−1=x2+x21−(2×x×x1)−1=[x2+x21−(2×x×x1)]−1=(x−x1)2−1=(x−x1)2−12=(x−x1+1)(x−x1−1).
Hence, Option 4 is the correct option.
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Factorise :
a - b - a3 + b3
x4+1x4−2x^4 + \dfrac{1}{x^4} - 2x4+x41−2 in the form of factors is :
(x−1x)2(x+1x+1)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} + 1\Big)^2(x−x1)2(x+x1+1)2
(x−1x)2(x+1x)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2(x−x1)2(x+x1)2
(x+1x)2(x−1x+1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x - \dfrac{1}{x} + 1\Big)^2(x+x1)2(x−x1+1)2
(x+1x)2(x+1x−1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} - 1\Big)^2(x+x1)2(x+x1−1)2
x2+14x2+1−7x−72xx^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x}x2+4x21+1−7x−2x7
9a2+19a2−2−12a+43a9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a}9a2+9a21−2−12a+3a4
