Advertisements
Advertisements
प्रश्न
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
पर्याय
196
289
324
400
Advertisements
उत्तर
We have to find the value of `{(23+2^2)^(2/3)+ (140- 19 )^(1/2) }^2`
`{(23+2^2)^(2/3)+ (140- 19 )^(1/2) }^2 = {(23+4)^(2/3)+ (121)^(1/2) }^2`
= `{(27)^(2/3)+ (121)^(1/2) }^2`
`={(3^3)^(2/3)+ (11^2)^(1/2) }^2`
`{(23+2^2)^(2/3)+ (140- 19 )^(1/2) }^2`= ` {3^(3 xx2/3) +11
^( 2xx 1/2)}^2`
` = {3^(3 xx2/3) +11^( 2xx 1/2)}^2`
= `{3^2 + 11}^2`
`⇒ {(23+2^2)^(2/3)+ (140- 19 )^(1/2) }^2 = {9+11}^2`
By using the identity `(a+b)^2 = a^2 +2ab +b^2` we get,
`= 9 xx 9 +2 xx 9 xx 11 + 11 xx 11`
`= 81 +198 +121`
`= 400`
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Prove that:
`(1/4)^-2-3xx8^(2/3)xx4^0+(9/16)^(-1/2)=16/3`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` find x and y.
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
If 24 × 42 =16x, then find the value of x.
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
Find:-
`32^(2/5)`
Simplify:
`11^(1/2)/11^(1/4)`
