Advertisements
Advertisements
प्रश्न
Evaluate the following:
`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`
Advertisements
उत्तर
`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`
= `(27/8)^(2/3) - (3)^2 - 1`
= `(3/2)^(3 xx 2/3) - 9 - 1`
= `(3/2)^2 - 10`
= `(9)/(4) - 10`
= `(9 - 40)/(4)`
= `(-31)/(4)`.
APPEARS IN
संबंधित प्रश्न
Find x, if : `( sqrt(3/5))^( x + 1) = 125/27`
If 5-P = 4-q = 20r, show that : `1/p + 1/q + 1/r = 0`
Evaluate : `[(-2/3)^-2]^3 xx (1/3)^-4 xx 3^-1 xx 1/6`
Solve : 3x-1× 52y-3 = 225.
If `((a^-1b^2 )/(a^2b^-4))^7 ÷ (( a^3b^-5)/(a^-2b^3))^-5 = a^x . b^y` , find x + y.
Evaluate : `4/(216)^(-2/3) + 1/(256)^(-3/4) + 2/(243)^(-1/5)`
Evaluate the following:
`(2^3 xx 3^5 xx 24^2)/(12^2 xx 18^3 xx 27)`
If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0
If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)3 = 27xyz
Prove the following:
`(x^("p"("q"-"r")))/(x^("q"("p"-"r"))) ÷ (x^"q"/x^"p")^"r"` = 1
