I like symmetry! In a third method substitute x + 2 = u 1/(u-2) + 1/(u-1) + 1/u + 1/(u+1) + 1/(u+2) = 0 After some multiplying out (u+2)u(u^2-1) + (u+1)u(u^2-4) + (u^2-1)(u^2-4) + (u-1)u(u^2-4) + (u-2)u(u^2-1) = 0 Grouping the first and fifth and the second and fourth terms 2u^2(u^2-1) + 2u^2(u^2-4) + (u^2-1)(u^2-4) = 0 2u^2(2u^2-5) + (u^2-1)(u^2-4) = 0 5u^4 - 15u^2 + 4 = 0 and the solution proceeds as Method 1
Yeah, I solved it the same way. This is the first method we should try when having the symmetry like that)) By the way the second method in the video uses the same trick with grouping of symmetrical terms but without substitution x + 2 = u though which makes it a way harder...
Substitute t=x+2, Then it becomes 1/(t-2) + 1/(t-1) + 1/t + 1/(t+1) + 1/(t+2) = 0 Now proceed the same as the second method, grouping and adding 1st and last terms together, and 2nd and 2nd last terms together. It becomes:- 2t/(t²-4) + 2t/(t²-1) = 1/t → 1/(t²-4) +1/(t²-1) = 1/2t² Put u = t² and solve for u.
First comment PIN PLEASE
I like symmetry! In a third method substitute
x + 2 = u
1/(u-2) + 1/(u-1) + 1/u + 1/(u+1) + 1/(u+2) = 0
After some multiplying out
(u+2)u(u^2-1) + (u+1)u(u^2-4) + (u^2-1)(u^2-4) + (u-1)u(u^2-4) + (u-2)u(u^2-1) = 0
Grouping the first and fifth and the second and fourth terms
2u^2(u^2-1) + 2u^2(u^2-4) + (u^2-1)(u^2-4) = 0
2u^2(2u^2-5) + (u^2-1)(u^2-4) = 0
5u^4 - 15u^2 + 4 = 0
and the solution proceeds as Method 1
Yeah, I solved it the same way. This is the first method we should try when having the symmetry like that))
By the way the second method in the video uses the same trick with grouping of symmetrical terms but without substitution x + 2 = u though which makes it a way harder...
Thanks for sharing!
@@SyberMath I solveded the same way substituting x+2=u.
The first method wasn’t as bad as as I thought it would be
Substitute t=x+2,
Then it becomes
1/(t-2) + 1/(t-1) + 1/t + 1/(t+1) + 1/(t+2) = 0
Now proceed the same as the second method, grouping and adding 1st and last terms together, and 2nd and 2nd last terms together. It becomes:-
2t/(t²-4) + 2t/(t²-1) = 1/t
→ 1/(t²-4) +1/(t²-1) = 1/2t²
Put u = t² and solve for u.
2t/(t²-4) + 2t/(t²-1) = -1/t
@@PeterNguyen-fo5hq oh thanks, silly mistake
Very nice💥💥💥
Thanks 🔥
Cool - I think the second method is definitely better!
Definitely!
🔥😎👍✌️👍🔥
Fresh
?
(x^2+3x+2+x^2+2x+x^2+x)/x(x+1)(x+2)=(-x-4-x-3)/(x+3)(x+4)..(3x^2+6x+2)(x+3)(x+4)=x(x+1)(x+2)(-2x-7)...(3x^2+6x+2)(x^2+7x+12)=(x^2+x)(-2x^2-11x-14)....3x^4+27x^3+80x^2+86x+24=-2x^4-13x^3-25x^2-14x...5x^4+40x^3+105x^2+100x+24=0...ah
1/x + 1/(x + 1) + 1/(x + 2) + 1/(x + 3) + 1/(x + 4) = 0
y = x + 2
1/(y - 2) + 1/(y - 1) + 1/y + 1/(y + 1) + 1/(y + 2) = 0
2y/(y^2 - 4) + 2y/(y^2 - 1) + 2y/(y^2) = 0
2y = 2(x + 2) cannot be zero
1/(y^2 - 4) + 1/(y^2 - 1) + 1/(y^2) = 0
z = y^2 - 2
1/(z - 2) + 1/(z + 1) + 1/(z + 2) = 0
2z/(z^2 - 4) + 1/(z + 1) = 0
3z^2 + 2z - 4 = 0
z = (-1 +/- ✓13)/3
y^2 = z + 2 = (5 +/- ✓13)/3
y = ✓[(5 +/- ✓13)/3], - [(5 +/- ✓13)/3]
x = y - 2 = - 2 + ✓[(5 +/- ✓13)/3], -2 - [(5 +/- ✓13)/3]
阴阳怪气！！！