We know that,

2+2=4

=4-\frac {9}{2}+\frac {9}{2}

=\sqrt {(4-\frac {9}{2})^2}+\frac {9}{2}

=\sqrt {16-2×4×(\frac {9}{2})+(\frac{9}{2})^2}+\frac {9}{2}

=\sqrt{16-36+(\frac {9}{2})^2}+\frac {9}{2}

=\sqrt{25-45+(\frac {9}{2})^2}+\frac {9}{2}

=\sqrt{(5)^2-2×5×\frac {9}{2}+(\frac {9}{2})^2}+\frac {9}{2}

=\sqrt{(5-\frac {9}{2})^2}+\frac {9}{2}

=5-\frac {9}{2}+\frac {9}{2}

=5

=\fbox {(2+2=5)Proved}

**But here is a mistake. Find it.**

I make here a writing mistake. But I edited it. Find now.

It is not always true that √(x^2)=x. It is applicable only when x is greater than or equal to 0. In your process, this law has been broken.

Brilliant access, this is the mistake. Good work fattah.

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\fbox{I have also thought like that.}

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We know that,

x^0=1 ; if, x^a=y^a then, x=y

So,

\implies {1=1}

\implies(1)^0=(2)^0

\implies{1=2} [x^a=x^b ,then , x=y]

**Finally 1=2**

1⁰=2⁰

–>\frac{1}{1}= \frac{2}{2}

–>\frac{1×2}{1}=\frac{2×2}{2}$

–>2=2

1⁰=2⁰

–>\frac{1}{1}=\frac{2}{2}

–>\frac{1×1}{1}=\frac{2×1}{2}$

–>1=1

1≠2

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x^a=y^a is applicable when a is not equal to 0. If a=0 everything will be 1 then.

Ok. Now,

Assume that, x=1 & y=2 & a=3

Then,

x^a=y^a

\implies{1^3=2^3}

\implies{1=2}

**But here is another **