Question

$\begin{gathered}\sf\ If\ \ abx^2= (a-b)^2(x+1)\\ \\ \sf\ then\ find\ the \ value\ of\ \\ \\ \sf\ x+\dfrac{4}{x^2}+\dfrac{4}{x}\end{gathered}$​

Answer 1

Correct Question :

If abx² = ( a - b)²( x + 1) , then find the value of

1 + 4/x^2 + 4/x

Solution :

Here , the starting expression given is that :

abx^2 = ( a - b)²( x + 1)

We need to simplify this to get terms with the denominators 1/x^2 and 1/x

Substituting those values we will obtain the required answer.

So , let's proceed

abx^² = ( a - b)²( x + 1)

Transferring the a and b terms to one side and the x terms to the other , divide the LHS by ( x + 1) and the RHS by ab

=> [ x²/( x + 1)] = ( a - b)²/ab

Let us invert this now ;

=> ( x + 1)/x^2 = ab/(a-b)²

Expanding the denominator in the LHS

=> 1/x + 1/x² = ab/(a-b)² .

We will use this in the value later .

Let us return to the value of what we need to find ;

=> 1 + 4/x^2 + 4/x

=> 1 + 4[ 1/x + 1/x^2]

Let us find the value of 4[ 1/x + 1/x^2 ] first

=> 4 [ ab]/( a - b)²

=> { 4ab }/{ a - b}²

4ab is a identity which we will convert in terms of square values to get this further simplified

4ab = ( a² + b²) - ( a² + b²) + 2ab + 2ab

=> [ a² + b² + 2ab ] - [ a² - 2ab + b²]

=> ( a + b)² - ( a - b)²

This finally becomes :

=> [ ( a + b)² - ( a - b)²]/[ ( a - b)²]

Let us divide the numerator by the denominator to further simplify this :

=> { [ a + b]/[a - b] } ² - 1

Required value :

=> 1 + { [ a + b]/[a - b] } ² - 1

=>{ [ a + b ]/[ a - b] } ²

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Correct Answer

{ [ a + b ]/[ a - b] } ²

Answer 2

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$abx^2=(a-b)^2(x+1)$

$\implies \dfrac{ab}{(a+b)^2}=\dfrac{x+1} {x^2}$

$\implies \dfrac{1}{x}+\dfrac{1} {x^2}=\dfrac{ab}{(a+b)^2}$

$\implies 1+\dfrac{4}{x}+\dfrac{4} {x^2}=1+\dfrac{4ab}{(a-b)^2}$

$\implies 1+\dfrac{4}{x}+\dfrac{4} {x^2}=\dfrac{(a-b)^2+4ab}{(a-b)^2}$

$\implies 1+\dfrac{4}{x}+\dfrac{4} {x^2}=\dfrac{(a-b)^2+(a+b)^2-(a-b)^2}{(a-b)^2}$

$\implies 1+\dfrac{4}{x}+\dfrac{4} {x^2}=\dfrac{(a+b)^2}{(a+b)^2}$

$\implies 1+\dfrac{4}{x}+\dfrac{4} {x^2}=\left(\dfrac{a+b}{a-b}\right)^2$

beech ka ek step khaa gyi mai..xd