Respuestas
SOLUCIÓN :
Racionalizar cada expresión :
a ) 4ab/∛x²y³z³b = 4ab/yz∛x²b *∛xb²/∛xb² = 4ab∛xb²/yzxb
= 4a∛xb² /yzx
b) h/(√x+h -√x )*( √x+h +√x )/( √x+h + √x ) = h*(√x+h +√x )/h =√x+h + √x
c ) m³n√x/ ⁵√2⁴m⁷n⁶x = m³n√x/mn⁵√2⁴m²nx = m²√x / ⁵√2⁴m²nx* ⁵√2m³n⁴x⁴ / ⁵√2m³n⁴x⁴ = m²√x*⁵√2m³n⁴x⁴ /2mnx = m*√x*⁵√2m³n⁴x⁴ / 2nx
d) √m+1 /( 1-√m+1 ) * ( 1+√m+1 ) /( 1+ √m+1 ) = ( √m+1 + m+1 )/( 1 -(m +1 )) = (√m+1 + m +1 )/-m = - ( √m+1 +m+1 )/m .
e) 3ab²/√ab⁵ = 3ab²/b²√ab = 3a/√ab*√ab/√ab= 3a√ab/ab = 3√ab /b
f) x/(√x²+1 -x )*( √x²+1 + x )/( √x²+1 + x ) = x * ( √x²+1 +x )
g) √1+x /√1-x *√1-x /√1-x = √1-x² / (1-x )
h) 2 / ( √(2+√2) - √2 ) * ( √( 2+√2) + √2 )/( √( 2+√2) +√2 )
= 2* ( √(2+√2) +√2 ) /√2 *√2/√2 = √2* (√( 2+√2) +√2 )
i) ( 3 -√a)/( √a -2 ) * ( √a +2 )/(√a +2 ) = ( 3√a +6 -(√a)²-2√a )/(a-4)
= ( √a +6-a)/(a-4 )
j) ( √x+1 + √x-1 )/( √x-1 +√x+1 )* ( √x-1 -√x+1 )/(√x-1 -√x+1 )= ( √(x²-1) -(√x+1 )²+ ( √x-1 )²-√(x²-1) / (x-1)-(x+1 ) = -2/-2 = 1
Respuesta:
a ) 4ab/∛x²y³z³b = 4ab/yz∛x²b *∛xb²/∛xb² = 4ab∛xb²/yzxb
= 4a∛xb² /yzx
b) h/(√x+h -√x )*( √x+h +√x )/( √x+h + √x ) = h*(√x+h +√x )/h =√x+h + √x
c ) m³n√x/ ⁵√2⁴m⁷n⁶x = m³n√x/mn⁵√2⁴m²nx = m²√x / ⁵√2⁴m²nx* ⁵√2m³n⁴x⁴ / ⁵√2m³n⁴x⁴ = m²√x*⁵√2m³n⁴x⁴ /2mnx = m*√x*⁵√2m³n⁴x⁴ / 2nx
d) √m+1 /( 1-√m+1 ) * ( 1+√m+1 ) /( 1+ √m+1 ) = ( √m+1 + m+1 )/( 1 -(m +1 )) = (√m+1 + m +1 )/-m = - ( √m+1 +m+1 )/m .
e) 3ab²/√ab⁵ = 3ab²/b²√ab = 3a/√ab*√ab/√ab= 3a√ab/ab = 3√ab /b
f) x/(√x²+1 -x )*( √x²+1 + x )/( √x²+1 + x ) = x * ( √x²+1 +x )
g) √1+x /√1-x *√1-x /√1-x = √1-x² / (1-x )
h) 2 / ( √(2+√2) - √2 ) * ( √( 2+√2) + √2 )/( √( 2+√2) +√2 )
= 2* ( √(2+√2) +√2 ) /√2 *√2/√2 = √2* (√( 2+√2) +√2 )
i) ( 3 -√a)/( √a -2 ) * ( √a +2 )/(√a +2 ) = ( 3√a +6 -(√a)²-2√a )/(a-4)
= ( √a +6-a)/(a-4 )
j) ( √x+1 + √x-1 )/( √x-1 +√x+1 )* ( √x-1 -√x+1 )/(√x-1 -√x+1 )= ( √(x²-1) -(√x+1 )²+ ( √x-1 )²-√(x²-1) / (x-1)-(x+1 ) = -2/-2 = 1v
Explicación paso a paso: