Question

1/(x²+1/x)^{4/3}) can be expanded by binomial themorem if

a)x<1

b)|x|<1

c)x>1

d)|x|>1

*Please give a detailed solution and irrelavent answer for gaining points will be reported immediatly!*

Answer 1

Answer:

sale

Step-by-step explanation:

sale

Answer 2

Answer:

Step-by-step explanation:

Given expression: 1/(x²+1/x)^{4/3}

Step 1: Simplify the denominator

(x²+1/x) = (x³+1)/x

Step 2: Substitute (x³+1)/x for (x²+1/x) in the expression

1/[(x³+1)/x]^{4/3}

Step 3: Move x^{4/3} inside the square brackets by multiplying and dividing by x^{4/3}

= 1/[(x³+1)^{4/3}/x^{4/3}]

= (x^{-4/3}*(x³+1)^{4/3})/1

Step 4: Apply the binomial theorem to expand (x³+1)^{4/3} if possible.

The binomial theorem is applicable if |x³|<1, i.e., if |x|<1.

Step 5: Expand (x³+1)^{4/3} using the binomial theorem

(x³+1)^{4/3} = 1 + (4/3)x³ + (4/3)*(1/3)^2 + ...

Using the formula for the binomial theorem:

(x³+1)^{4/3} = ∑[k=0 to 4] (4/3 choose k) (x³)^k (1)^{4-k}

= 1 + (4/3)x³ + (16/27)x^6 + (64/81)x^9 + (256/243)x^12

Step 6: Substitute the expansion back into the original expression

= (x^{-4/3})*(1 + (4/3)x³ + (16/27)x^6 + (64/81)x^9 + (256/243)x^12)

Step 7: Simplify the expression by multiplying x^{-4/3} with each term

= x^{-4} + (4/3)x^{-1} + (16/27)x^{2/3} + (64/81)x^{5/3} + (256/243)x^4

Therefore, the expansion of the expression 1/(x²+1/x)^{4/3} using the binomial theorem is possible if |x|<1, and the final expanded expression is given by:

x^{-4} + (4/3)x^{-1} + (16/27)x^{2/3} + (64/81)x^{5/3} + (256/243)x^

The answer is (b) |x|<1, as we saw in the detailed solution that the binomial theorem can be applied if |x³|<1, which is equivalent to |x|<1.