Find the square of:
x+1x−1x +\dfrac{1}{x}- 1x+x1−1
3 Likes
(x+1x−1)2(x +\dfrac{1}{x}- 1)^2(x+x1−1)2
Using the formula
(∵ (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz)
=(x)2+(1x)2+(1)2+2×x×(1x)−2×(1x)×1−2×1×x=x2+(1x2)+1+(2xx)−(2x)−2x=x2+(1x2)+1+2−(2x)−2x=x2+(1x2)+3−(2x)−2x= (x)^2 + \Big(\dfrac{1}{x}\Big)^2 + (1)^2 + 2 \times x \times \Big(\dfrac{1}{x}\Big) - 2 \times \Big(\dfrac{1}{x}\Big) \times 1 - 2 \times 1 \times x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + \Big(\dfrac{2x}{x}\Big) - \Big(\dfrac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + 2 - \Big(\dfrac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 3 - \Big(\dfrac{2}{x}\Big) - 2x=(x)2+(x1)2+(1)2+2×x×(x1)−2×(x1)×1−2×1×x=x2+(x21)+1+(x2x)−(x2)−2x=x2+(x21)+1+2−(x2)−2x=x2+(x21)+3−(x2)−2x
Hence, (x+1x−1)2(x +\dfrac{1}{x}- 1)^2(x+x1−1)2 = x2 + (1x2)\Big(\dfrac{1}{x^2}\Big)(x21) + 3 - (2x)\Big(\dfrac{2}{x}\Big)(x2) - 2x
Answered By
1 Like
5−x+2x5 - x +\dfrac{2}{x}5−x+x2
2x - 3y + z
Evaluate using expansion of (a + b)2 or (a - b)2 :
(i) (208)2
(ii) (92)2
(iii) (9.4)2
(iv) (20.7)2
Expand:
(2a + b)3
