Question

Q1. The value of {(2/3)22 is is​

Answer 1

QUESTION

If x=2+2

2/3+2 1/3 , then the value of x 3 −6x 2+6x is

ANSWER

Given, x=2+2

Given, x=2+2 2/3

Given, x=2+2 2/3 +2

Given, x=2+2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2)

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)⇒x

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)⇒x 3

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)⇒x 3 −6x

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)⇒x 3 −6x 2

Given, x=2+2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 ⇒(x−2)=2 2/3 +2 1/3 On cubing both sides, we get(x−2) 3 =2 2 +2+3⋅2 2/3 ⋅2 1/3 (2 2/3 +2 1/3 )⇒x 3 −6x 2 +12x−8=4+2+3×2×(x−2)⇒x 3 −6x 2 +6x=2.

Answer 2

Answer:

22multiple 2+3= 77

your ans is 77 value OK

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