Question

square root of x+y+z+√xy+√yz+√xz​

Answer 1

Answer:

GIVEN: xyz + xy + xz + yz + x + y + z = 384

TO FIND: x+y+z

This cyclic expression can be factorized first. As we notice individual power of all 3 variables in every term , is not more than 1. So there are chances that its each factor contains just one of the 3 variables…

xyz +xy + yz + xz +x + y + z = 384

=> xy( z+1) + z( x + y) + (x+y) + z = 384

=> xy(z+1) +( x+y)(z+1) + z = 384

=> xy(z+1) + (x+y)(z+1) +( z+1) = 384 +1 (added 1 to both the sides)

=> (z+1) ( xy + x+y + 1) = 385

=> ( z+1) {x(y+1) + 1( y+1)} = 385

=> ( z+1)(y+1)( x+1) = 5 x 7 x 11

=> (x+1), (y+1), (z+1) hold either of these values. 5 or 7 or 11

=> x, y, z hold either of the values 4, 6 , or 10

Hence, x+y+ z = 4+6+10= 20 ●●●●●●ANS

Answer 2

QUESTION

square root of x+y+z+2√xy+2√yz+2√xz

WE KNOW THAT

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

SOLUTION

x + y + z + √xy + √yz + √xz

(√x)² + (√y)² + (√z)² + 2(√xy + √yz + √xz)

(√x + √y + √z)²

Square Root of (√x + √y + √z)² is (√x + √y + √z)