Advertisements
Advertisements
प्रश्न
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
उत्तर
We have to prove that `sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Let x = `sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)`
`=sqrt(1/2^2)+((0.01xx100)/(1xx100))^(-1/2)-(3^3)^(2/3)`
`=1/2+1/(100)^(-1/2)-3^(3xx2/3)`
`=1/2+1/(1/100^(1/2))-3^2`
`=1/2+1/(1/(10xx10)^(1/2))-3^2`
`=1/2+1/(1/10^(2xx1/2))-3^2`
`=1/2+1/(1/10)-3^2`
`=1/2+1xx10/1-3xx3`
`=1/2+10-9`
`=3/2`
Hence, `sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
APPEARS IN
संबंधित प्रश्न
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`5^(2x+3)=1`
If `x=2^(1/3)+2^(2/3),` Show that x3 - 6x = 6
If `5^(3x)=125` and `10^y=0.001,` find x and y.
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
Find:-
`32^(1/5)`
