Advertisements
Advertisements
प्रश्न
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Advertisements
उत्तर
It is given that `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1)`
`thereforea^(q-r)b^(r-p)c^(p-q)`
`=(xy^(p-1))^(q-r)(xy^(q-1))^(r-p)(xy^(r-1))^(p-q)`
`=x^((q-r))y^((p-1)(q-r))x^((r-p))y^((r-p)(q-1))x^((p-q))y^((p-q)(r-1))`
`=x^((q-r))x^((r-p))x^((p-q))y^((p-1)(q-r))y^((r-p)(q-1))y^((p-q)(r-1))`
`=x^((q-r)+(r-p)+(p-q))y^((p-1)(q-r)+(r-p)(q-1)+(p-q)(r-1))`
`=x^(q-r+r-p+p-q)y^(pq-q-pr+r+rq-r-pq+p+pr-p-qr+q)`
`=x^0y^0`
= 1
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Prove that:
`(a^-1+b^-1)^-1=(ab)/(a+b)`
Simplify:
`root3((343)^-2)`
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If x-2 = 64, then x1/3+x0 =
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
If (16)2x+3 =(64)x+3, then 42x-2 =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
