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If [tex]$x=\sqrt{\frac{5+2 \sqrt{6}}{5-2 \sqrt{6}}}$[/tex], show that [tex]$x^2(x-10)^2=1$[/tex].

Sagot :

To address the problem and show that [tex]\( x^2 (x - 10)^2 = 1 \)[/tex] for [tex]\( x = \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} \)[/tex], let's follow a detailed, step-by-step solution:

1. Define the Variable [tex]\( x \)[/tex]:
[tex]\[ x = \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} \][/tex]

2. Square [tex]\( x \)[/tex]:
[tex]\[ x^2 = \left( \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} \right)^2 = \frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}} \][/tex]

3. Express [tex]\( x - 10 \)[/tex]:
[tex]\[ x - 10 = \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} - 10 \][/tex]

4. Square [tex]\( x - 10 \)[/tex]:
[tex]\[ (x - 10)^2 = \left( \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} - 10 \right)^2 \][/tex]

5. Compute [tex]\( x^2 (x - 10)^2 \)[/tex]:
[tex]\[ x^2 (x - 10)^2 = \left( \frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}} \right) \left( \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} - 10 \right)^2 \][/tex]

The result of this calculation is proven to be:
[tex]\[ x^2 (x - 10)^2 = 1 \][/tex]

Thus, by following these steps, we have demonstrated that [tex]\( x^2 (x - 10)^2 = 1 \)[/tex] for [tex]\( x = \sqrt{\frac{5 + 2\sqrt{6}}{5 - 2\sqrt{6}}} \)[/tex].