prove that 1/1+a^x-y+1/1+a^y-x =1
Answers
Answer:
Let (1+x+1y)=t , then yt=1+y+xy
Consider, (1+y+1z) . Multiply and divide by xz .
xz+xyz+xxz=y(1+x+1y)=yt …..using xyz=1
Consider, (1+x+1y) . Multiply and divide by yz .
yz+xyz+zyz=(1+z+1x)yz
i.e., yztyz=(1+z+1x)yz
i.e., yzt=(1+z+1x)
i.e., (1+z+1x)=tx …..using xyz=1
So the LHS of the problem becomes:
1t+1yt+xt
=1+y+xyyt=ytyt=1
Riya Rohilla
Answered February 4, 2018Siddhant Grover
Muthusamy Piramanayagam ( முத்துசாமி பிரமநாயகம்)
Updated January 7, 2020
Let P = ( 1+ x + 1/y ) = ( 1+ x + xz) since xyz = 1————-(1)
Let Q= ( 1+ y+ 1/z) = ( 1+ y + xy ) ————-(2)
Let R= ( 1+ z+ 1/x) = ( 1+ x + yz) —————(3)
Py = ( y+xy + 1) = Q , ————(4) ,
Pyz = (zy + 1 + z) =R ————-(5)
( 1/P + 1/ Q + 1/ R ) = (1/P) ( 1 + 1/y + 1/ zy)
( 1/P + 1/ Q + 1/ R ) = (1/P) ( 1 +xz+ x)
= (1/P) *P = 1. And the proof.
Tanu Shyam Majumder
Answered September 21, 2018
xyz=1. Let x=a/b,y=b/c,z=c/a
1+x+y^(-1)=1+a/b+c/b=(a+b+c)/b
1+y+z^(-1)=1+b/c+a/c=a+b+c/c
1+z+x^(-1)=1+c/a+b/a=a+b+c/a
(1+x+y^(-1))^(-1)+(1+y+z^(-1))^(-1)+ (1+z+x^(-1))^(-1)=b/a+b+c + c/a+b+c + a/a+b+c=a+b+c/a+b+c=1(answer)
Harsh Dwivedi
Answered January 17, 2020
11+x+1y+11+y+1z+11+z+1xA key Note is provided to us thatxyz=1yy+xy+1+zz+zy+1+xx+xz+1x=1yz,xz=1z,xy=1zyy+1z+1+zy+zy+1+x1yz+1y+1yzz+yz+1+zy+zy+1+xyzz+yz+1z+yz+1z+yz+1=1Hence,proved■
Space Mafia
Answered January 21, 2018
Mayuree
Answered January 21, 2018
=1(1+x+xz)+1(1+(xz)−1+z−1)+1(1+z+x−1) =1(1+x+xz)+1(1+(xz)−1+z−1)+1(1+z+x−1)
=1(1+x+xz)+xz(xz+1+x)+x(x+xz+1) =1(1+x+xz)+xz(xz+1+x)+x(x+xz+1)
=1(1+x+xz)+xz(1+x+xz)+x(1+x+xz) =1(1+x+xz)+xz(1+x+xz)+x(1+x+xz)
=1+x+xz(1+x+xz) =1+x+xz(1+x+xz)
=1
Vishal Shukla
Answered April 5, 2018
Substitute z=1/xy in LHS.
And solve for LHS you will get it equal to RHS.
LHS= (y/(1+xy+y))+(1/(1+y+xy))+(xy/(1+y+xy))
= (1+xy+y)/(1+xy+y)
=1
Hope it may help you.
Amit Shukla
Answered December 2, 2017
(1+x+y^-1)^-1+(1+y+z^-1)^-1+(1+z+x^-1)^-1
={y/(y+xy+1)}+{z/(z+yz+1)}+{x/(x+zx+1)}
Putting y=1/zx
1/(1+x+zx)+zx/(zx+1+x)+x/(x+zx+1)
= (1+x+zx)/(1+x+zx)
=1