Advertisements
Advertisements
प्रश्न
Show that:
`(("x"^"a")/"x"^(-"b"))^("a"-"b").(("x"^"b")/"x"^(-"c"))^("b"-"c").(("x"^"c")/("x"^(-"a")))^("c"-"a")=1`
उत्तर
`"L" . "H" ."S".`
`(("x"^"a")/"x"^(-"b"))^("a"-"b").(("x"^"b")/"x"^(-"c"))^("b"-"c").(("x"^"c")/("x"^(-"a")))^("c"-"a")=1`
`=("x"^("a"+"b"))^("a"-"b").("x"^("b"+"c"))^("b"-"c").("x"^("c"+"a"))^("c"-"a")`
`="x"^("a"^2-"b"^2)."x"^("b"^2-"c"^2)."x"^("c"^2-"a"^2)`
`="x"^("a"^2-"b"^2+"b"^2-"c"^2+"c"^2-"a"^2`
`="x"^0`
`=1 ="R"."H"."S".`
APPEARS IN
संबंधित प्रश्न
Compute:
`1^8xx3^0xx5^3xx2^2`
Compute:
`(4^7)^2xx(4^-3)^4`
Compute:
`(56/28)^0÷(2/5)^3xx16/25`
Compute:
`(-1/3)^4÷(-1/3)^8xx(-1/3)^5`
Compute:
`(125)^(-2/3)÷(8)^(2/3)`
Evaluate:
`4"x"^0`
Evaluate:
`[(10^3)^0]^5`
Simplify:
`"x"^10"y"^6÷"x"^3"y"^-2`
Simplify:
`(36"x"^2)^(1/2)`
Evaluate:
`("a"^(2"n"+1)xx"a"^((2"n"+1)(2"n"-1)))/("a"^("n"(4"n"-1))xx("a"^2)^(2"n"+3)`