Question

Simplify: √(1+x) /√(1+x) - √(1 -x) - 1-x /√(1 -x) + x -1
$simplify : \frac{ \sqrt{ 1 + x}}{ \sqrt{1 + x} - \sqrt{1 - x} } - \frac{1 - x}{ \sqrt{1 - {x}^{2} } + x - 1 }$
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Answer 1

Step-by-step explanation:

Given that: [√(1+x)/{√(1+x) - √(1-x)}] - [(1-x)/{√(1-x²) + x -1}]

Here, we see that in first fraction the denominator is √(1+x)-√(1-x) , as we know that the rational factor √(a+b)-√(a-b) is √(a+b)+(a-b). Therefore, the rationalising factor of √(1+x)-√(1-x) is √(1+x)+√(1-x). On rationalising the denominator them.

= [√(1+x)/{√(1+x)-√(1-x)}] * [{√(1+x)+√(1-x)}/{√(1+x)+√(1-x)}] - [(1-x)/{√(1-x²) + x -1}]

= [{√(1+x)(√(1+x)+√(1-x)}/{√(1+x)-√(1-x) - √(1+x)+√(1-x)}] - [(1-x)/{√(1-x²) + x -1}]

Now, comparing the first fraction denominator with (a-b)(a+b), we get

• a = √(1+x)
• b = √(1-x)
• a = √(1+x)
• b = √(1-x)

Using identity (a-b)(a+b) = a² - b², we get

= [{√(1+x)(√(1+x)+√(1-x)}/{√(1+x)² - √(1-x)²}] - [(1-x)/{√(1-x²) + x -1}]

= [{√(1+x)(√(1+x)+√(1-x)}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]

Now, multiply the numerator on both brackets.

= [{√(1+x) * √(1+x) + √(1+x) * √(1-x)}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]

Comparing the first fraction numerator with (a+b)(a+b) , we get

• a = √1
• b = √x

Using identity (a+b)(a+b) = (a+b)², we get

= [{√(1+x)²+ √(1+x) * √((1+x)(1-x))}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]

Cancel out the first fraction denominator numbers 1 and -1 to get 0.

= [{(1+x)+√(1-x²)}/(x+x+0)] - [(1-x)/{√(1-x²) + x -1}]

= [{(1+x)+√(1-x²)}/(x+x)] - [(1-x)/{√(1-x²) + x -1}]

= [{(1+x)+√(1-x²)}/2x] - [(1-x)/{√(1-x²) + x -1}]

= [{1+x+√(1-x²)}{√(1-x²)x-1}-{(1-x)(2x)}]/[2x{√(1-x²)+x-1}]

= [√(1-x²) + x-1 + x√(1-x²) + x² - x + {√(1-x²)}² + x√(1-x²)-√(1-x²) - 2x + 2x²]/[2x{√(1-x²)+x-1}]

= {(√(1-x²) 2x -1 + 3x² + 2x√(1-x²) + 1-x² - √(1-x²)}/[2x{√(1-x²)+x-1}]

Cancel out √(1-x) and -√(1-x) in numerator.

= {-2x - 1 + 2x√(1-x²) + 3x² + 1 - x²}/[2x{√(1-x²)}+x-1}]

Cancel out -1 and 1 in numerator to get 0.

= [2x√(1-x²) - 2x + 2x²}/[2x{√(1-x²)+x-1}]

= {2x{√(1-x²)} + x - 1}]/[2x{√(1-x²)}+x-1}

= 1

Answer: Hence, the simplified form of the expression [√(1+x)/{√(1+x) - √(1-x)}] - [(1-x)/{√(1-x²) + x -1}] is 1.

Please let me know if you have any other questions.

Answer 2

The simplified expression of $\frac{\sqrt{1+x}}{\sqrt{1+x} - \sqrt{1-x}} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$ is 1

The expression is given as:

$\frac{\sqrt{1+x}}{\sqrt{1+x} - \sqrt{1-x}} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$

Rationalize the first fraction

$\frac{\sqrt{1+x}}{\sqrt{1+x} - \sqrt{1-x}} * \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$

Expand

$\frac{\sqrt{1+x}(\sqrt{1 + x} + \sqrt{1 - x})}{1+x - 1+x} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$

$\frac{\sqrt{1+x}(\sqrt{1 + x} + \sqrt{1 - x})}{2x} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$

Further, expand

$\frac{1+x + \sqrt{1 - x^2}}{2x} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$

Take LCM

$\frac{(1+x+\sqrt{1-x\²}(\sqrt{1-x\²} + x-1) - (1 - x)2x}{2x(\sqrt{1 -x^2} + x + 1)}$

Expand the expression

$\frac{\sqrt{1-x\²} + x-1 + x\sqrt{1-x\²} + x\² - x + (\sqrt{1-x\²})\² + x\sqrt{1-x\²}-\sqrt{1-x\²} - 2x + 2x\²}{2x(\sqrt{1-x\²}+x-1)}$

Simplify the expression

$\frac{(\sqrt{1-x\²}) 2x -1 + 3x\² + 2x\sqrt{1-x\²} + 1-x\² - \sqrt{1-x\²}}{2x(\sqrt{1-x\²}+x-1)}$

Cancel out the common factor in the fraction

$\frac{-2x - 1 + 2x\sqrt{1-x\²} + 3x\² + 1 - x\²}{2x(\sqrt{(1-x\²)}+x-1}$

Evaluate the like terms

$\frac{-2x + 2x\sqrt{1-x\²} + 2x\²}{2x(\sqrt{(1-x\²)}+x-1}$

Factor out 2x

$\frac{2x(\sqrt{(1-x\²)}+x-1}{2x(\sqrt{(1-x\²)}+x-1}$

Divide the fraction

1

Hence, the simplified expression of $\frac{\sqrt{1+x}}{\sqrt{1+x} - \sqrt{1-x}} -\frac{1-x}{\sqrt{1-x\² }+ x -1}$ is 1

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