If a=1a−7a = \dfrac{1}{a - 7}a=a−71, then a−1aa - \dfrac{1}{a}a−a1 =
0
1
7
17\dfrac{1}{7}71
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Given,
⇒a=1a−7⇒a×(a−7)=1⇒a−7=1a⇒a−1a=7\Rightarrow a = \dfrac{1}{a - 7} \\[1em] \Rightarrow a \times (a-7) = 1 \\[1em] \Rightarrow a - 7 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 7⇒a=a−71⇒a×(a−7)=1⇒a−7=a1⇒a−a1=7
Hence, option 3 is correct option.
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If ab+ba=1\dfrac{a}{b} + \dfrac{b}{a} = 1ba+ab=1 (a, b ≠ 0), then a3 + b3 =
2
8
If p+1p=x and p−1p=yp + \dfrac{1}{p} = x\text{ and }p - \dfrac{1}{p} = yp+p1=x and p−p1=y, then the relation between x and y is :
x2 = y2
x2 + y2 = 4
x2 - y2 = 4
xy = 2
2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51 =
1.44
0.44
If (x+1x)2=3\Big(x + \dfrac{1}{x}\Big)^2 = 3(x+x1)2=3, then x3+1x3x^3 + \dfrac{1}{x^3}x3+x31 =
9
3
