Advertisements
Advertisements
प्रश्न
Prove that:
`(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
Advertisements
उत्तर
We have to prove `(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
`(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=1^-2/4^-2-3xx2^(3xx2/3)xx4^0+3^(2xx-1/2)/2^(4xx-1/2)`
`=1/2^(2xx-2)-3xx2^2xx4^0+3^-1/2^-2`
`=1/2^-4-3xx2^2xx4^0+3^-1/2^-2`
`=1/(1/2^4)-3xx2^2xx4^0+(1/3)/(1/2^2)`
`=1xx2^4/1-3xx2^2xx1+1/3xx2^2/1`
`=16/1-12/1+4/3`
`=16/3`
Hence, `(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Simplify:
`root5((32)^-3)`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
(256)0.16 × (256)0.09
If o <y <x, which statement must be true?
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
Simplify:
`(3/5)^4 (8/5)^-12 (32/5)^6`
