NK
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2019 March 13
NK
and Guide
NK
Wtf Pura likh ke de rha kya?
R
simply the first eq as much as possible and you get
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
R
simply the first eq as much as possible and you get
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
R
p = 3, q =2
This I got. Did you get in that form tho ?
R
simply the first eq as much as possible and you get
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
.
R
simply the first eq as much as possible and you get
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
Wow. Thanks. Lemme try while following this
R
Chimera V6r2 emergency update is out!
Changelog:
Fixed KLapse
Update now if you were on V6
https://drive.google.com/file/d/1bOiI_pueqYZ-GSgW-5M51HKu-XLowEhh/
Changelog:
Fixed KLapse
Update now if you were on V6
https://drive.google.com/file/d/1bOiI_pueqYZ-GSgW-5M51HKu-XLowEhh/
/pin
R
Sorry unpinned by mistake
R
simply the first eq as much as possible and you get
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
1 - [c * (ab+1)]}/[b * (ac + 1)]
Ignore 1 for now and just solve the right side of the expression
solve ab+1/ac+1 for starters
ab + 1 = 6x^2 - x - 1
ac + 1 = 2x^2 - x
take x common in 2x^2 - x
6x^2 - x - 1 / x(2x - 1)
Use long division method(or whatever you know) to divide (6x^2 - x - 1) by (2x-1) to get 3x + 1
finally, right side of the eq =
[(x-1)(3x+1)]/[(3x-2)x]
Bring in the 1 now and you get
1 - [(x-1)(3x+1)]/x(3x-2)
= [3x^2 - 2x - 3x^2 + 2x + 1]/[3x^2 - 2x]
= 1/(3x^2 - 2x)
Comparing with given expression we get
p = 3, q = 2
You pro af
R
But that's 11th grade maths. So not big deal for you ig
NK
But that's 11th grade maths. So not big deal for you ig
No
R
But that's 11th grade maths. So not big deal for you ig
Dunno wut chapter
R
This not 11th
NK
11th grade ka nahi hai
R
9th probably
R
ID:719510732
11th grade ka nahi hai
Then ?