is there any genius to prove
that 1=2
LokeshSanthakumar:
you first prove 5=4 then i solve your problem
okk wait
Answers
Answered by
0
1) a = b 1) Given
2) a2 = ab 2) Multiply both sides by a
3) a2-b2 = ab-b2 3) Subtract b2 from both sides
4) (a+b)(a-b) = b(a-b)4) Factor both sides
5) (a+b) = b 5) Divide both sides by (a-b)
6) a+a = a 6) Substitute a for b
7) 2a = a 7) Addition
8) 2 = 1 8) Divide both sides by a
2) a2 = ab 2) Multiply both sides by a
3) a2-b2 = ab-b2 3) Subtract b2 from both sides
4) (a+b)(a-b) = b(a-b)4) Factor both sides
5) (a+b) = b 5) Divide both sides by (a-b)
6) a+a = a 6) Substitute a for b
7) 2a = a 7) Addition
8) 2 = 1 8) Divide both sides by a
then you solve
ADITYA0721 What's wrong in this answer?
he is *****
x^2-x^2=(x+x)(x-x)=x(x-x) cancelling x-x on both sides , therefore 2x=x hence 2=1 or 1=2 simple method
now who is *****
????
nothing(^^)(^^)
use identity and factorisation each time
a^2-b^2
if a=b=x then solve
Answered by
1
●a=c (GIVEN)
●a^2=ac (Multiplying both side by a)
●a^2 - c^2=ac-c^2 (Subtracting both side by c^2)
●(a+c) (a-c) = c(a-c)
●(a+c)=c (Dividing both side by a-c)
●c+c=c (substitute c for a)
●2c=c (addition)
●2=1 (Dividing both side by c)
●a^2=ac (Multiplying both side by a)
●a^2 - c^2=ac-c^2 (Subtracting both side by c^2)
●(a+c) (a-c) = c(a-c)
●(a+c)=c (Dividing both side by a-c)
●c+c=c (substitute c for a)
●2c=c (addition)
●2=1 (Dividing both side by c)
this is right
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