Question

root 1+x/root 1+x –root1–x – root 1–x /root 1–x^2 + x–1​

Answer 1

Step-by-step explanation:

$x = 9/13 or x = 4/13</p><p></p><p>Step-by-step explanation:</p><p></p><p>Here, the given expression,</p><p></p><p>\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}1−xx+x1−x=613</p><p></p><p>\frac{\sqrt{x}}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{\sqrt{x}}=\frac{13}{6}1−xx+x1−x=613</p><p></p><p>\frac{x+1-x}{\sqrt{x(1-x)}}=\frac{13}{6}x(1−x)x+1−x=613</p><p></p><p>\frac{1}{\sqrt{x(1-x)}}=\frac{13}{6}x(1−x)1=613</p><p></p><p>6=13[\sqrt{x(1-x)}]6=13[x(1−x)]</p><p></p><p>By squaring both sides,</p><p></p><p>36=169[x(1-x)]36=169[x(1−x)]</p><p></p><p>36=169x-169x^236=169x−169x2</p><p></p><p>169x^2-169x+36=0169x2−169x+36=0</p><p></p><p>By splitting the middle term,</p><p></p><p>169x^2-117x-52x+36=0169x2−117x−52x+36=0</p><p></p><p>13x(13x-9)-4(13x-9)=013x(13x−9)−4(13x−9)=0</p><p></p><p>(13x-9)(13x-4)=0(13x−9)(13x−4)=0</p><p></p><p>\implies 13x - 9 = 0\text{ or }13x - 4 = 0⟹13x−9=0 or 13x−4=0</p><p></p><p>\implies x = \frac{9}{13}\text{ or }x = \frac{4}{13}⟹x=139 or x=134</p><p></p><p>Which is the required solution.</p><p></p><p>$

Answer 2

Answer:

Here's your answer .............