Advertisements
Advertisements
प्रश्न
Simplify:
`("a"^(2"n"+3)."a"^((2"n"+1)("n"+2)))/(("a"^3)^(2"n"+1)."a"^("n"(2"n"+1)`
बेरीज
Advertisements
उत्तर
`("a"^(2"n"+3)."a"^((2"n"+1)("n"+2)))/(("a"^3)^(2"n"+1)."a"^("n"(2"n"+1))`
Given expression`=("a"^(2"n"+3)."a"^((2"n"^2+4"n"+"n"+2)))/("a"^(6"n"+3)."a"^(2"n"^2+"n"`
`="a"^(2"n"+3+2"n"^2+5"n"+2)/"a"^(6"n"+3+2"n"^2+"n")`
`="a"^(2"n"^2+7"n"+5)/"a"^(2"n"^2+7"n"+3`
`= a^(2n^2 +7n + 5 - 2n^2 - 7n - 3 )`
`= a^2`
shaalaa.com
More About Exponents
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
Compute:
`1^8xx3^0xx5^3xx2^2`
Compute:
`(2^-9÷2^-11)^3`
Compute:
`(125)^(-2/3)÷(8)^(2/3)`
Compute:
`(-3)^4-(root(4)(3))^0xx(-2)^5÷(64)^(2/3)`
Evaluate:
`8^0+4^0+2^0`
Evaluate:
`4"x"^0`
Evaluate:
`[(10^3)^0]^5`
Evaluate:
`(7"x"^0)^2`
Simplify:
`(2"x"^2"y"^-3)^-2`
Simplify and express as positive indice:
`("a"^(-2)"b")^(1/2)xx("a""b"^-3)^(1/3)`
