Forwarded from Questions collection (Yogesh)
11a+2=12b+4=13c+6
12b=11a-2
11*10-2=12b. B=9. A=10 number 112. Reminder 8 by 13
Will give solutions a will change
According to B's cofficient
B will according to a's cofficient
Same. B=9+11. A=10+12
B=9+11+11.... A=10+12+12
Next number 12*20+4=244 reminder 10 by 13
8,10,12,1,3,5,7,9,11,13,2,4,6
Reminder will be 6 after 11
12*(20+11*11)+4
12*(141)+4=1696
Number 1696
6 reminder will repeat after every 13 th Operation
Next. 12*(141+13*11)+4
12*284+4= 3412
Which is 3412-1696=1716
1696+2*1716=5128
1696+5*1716=10276
12b=11a-2
11*10-2=12b. B=9. A=10 number 112. Reminder 8 by 13
Will give solutions a will change
According to B's cofficient
B will according to a's cofficient
Same. B=9+11. A=10+12
B=9+11+11.... A=10+12+12
Next number 12*20+4=244 reminder 10 by 13
8,10,12,1,3,5,7,9,11,13,2,4,6
Reminder will be 6 after 11
12*(20+11*11)+4
12*(141)+4=1696
Number 1696
6 reminder will repeat after every 13 th Operation
Next. 12*(141+13*11)+4
12*284+4= 3412
Which is 3412-1696=1716
1696+2*1716=5128
1696+5*1716=10276
Forwarded from Questions collection (Yogesh)
R√10+R+1=√10
R=√10-1/√10+1
r√5+r+1=√5
r=√5-1/√5+1
r/R=(√5-1)*(√10+1)/(√5+1)(√10-1)
R=√10-1/√10+1
r√5+r+1=√5
r=√5-1/√5+1
r/R=(√5-1)*(√10+1)/(√5+1)(√10-1)
Forwarded from Math World (Azizakhon Turakhujaeva)
Forwarded from Math World (Azizakhon Turakhujaeva)
Forwarded from Deleted Account
x=3+2sqrt(2)
1/x=3-2sqrt2
x+1/x=6
x^2+1/x^2=(x+1/x)^2-2=36-2=34
1/x=3-2sqrt2
x+1/x=6
x^2+1/x^2=(x+1/x)^2-2=36-2=34
Forwarded from Math World (Azizakhon Turakhujaeva)
(80+48sqrt(3))^1/3=k+sqrt(3)b
80+48sqrt(3)=
k^3+3sqrt(3)k^2b+9b^2k+3sqrt(3)b
80+48sqrt(3)=k^3+9b^2k+3sqrt(3)(k^2b+b)
80=k^3+9b^2k
16=k^2b+b
b=2, k=2
=> (80+48sqrt(3))^1/3=2+2sqrt(3)
80+48sqrt(3)=
k^3+3sqrt(3)k^2b+9b^2k+3sqrt(3)b
80+48sqrt(3)=k^3+9b^2k+3sqrt(3)(k^2b+b)
80=k^3+9b^2k
16=k^2b+b
b=2, k=2
=> (80+48sqrt(3))^1/3=2+2sqrt(3)
Forwarded from Yogesh
3/1*3*5. 3 se 3 cut. Bacha 1/3*5
Same
1/1*5+1/3*7+1/5*9+1/7*11+...
Two series
(1/1*5+1/5*9+......)+(1/3*7+1/7*11+........)
1/4of (1-1/infinity)+. 1/4of (1/3-1/infinity)
1/4+1/12=1/3
Same
1/1*5+1/3*7+1/5*9+1/7*11+...
Two series
(1/1*5+1/5*9+......)+(1/3*7+1/7*11+........)
1/4of (1-1/infinity)+. 1/4of (1/3-1/infinity)
1/4+1/12=1/3
Forwarded from Deleted Account
(a+b+4){ab+4(a+b)}-4ab=0
{ab(a+b+4)+4(a+b)(a+b+4)} -4ab=0
ab(a+b)+4ab+4(a+b)²+16(a+b)-4ab=0
=(a+b){ab+4(a+b)+16}=0
(a+b){a(b+4)+4(b+4)}=0
(a+b)(a+4)(b+4)=0
Now
a≠-4,b≠-4
So
a+b=0
Now put (a+b=0) in the question
=[1/4^117 - 1/2²³⁴]
=[1/2²³⁴ - 1/2²³⁴] =0
{ab(a+b+4)+4(a+b)(a+b+4)} -4ab=0
ab(a+b)+4ab+4(a+b)²+16(a+b)-4ab=0
=(a+b){ab+4(a+b)+16}=0
(a+b){a(b+4)+4(b+4)}=0
(a+b)(a+4)(b+4)=0
Now
a≠-4,b≠-4
So
a+b=0
Now put (a+b=0) in the question
=[1/4^117 - 1/2²³⁴]
=[1/2²³⁴ - 1/2²³⁴] =0
Forwarded from Yogesh
Xy=4
(1,4) (2,2)
(1,4,2) not possible
(1,4,3) nope
(1,1,1) possible which is equliteral
(2,2,1). 4<5. Acute angle
(2,2,3). 9>8 possible
So z can be 3
(1,4) (2,2)
(1,4,2) not possible
(1,4,3) nope
(1,1,1) possible which is equliteral
(2,2,1). 4<5. Acute angle
(2,2,3). 9>8 possible
So z can be 3