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.
\fbox{I have also thought like that.}
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
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 
1^3=2^3 But How???
1³=2³
–>1=8
যেহেতু,1≠8
সুতরাং,1³≠2³