Knowing that a and b are both negative real numbers, and 1 a 1 b 1 a b 0 then what is the value of b

1\a+1\b+1\(a-b)=0

There are all points. b(a-b)+a(a-b)+ab] ab(a-b)=0, so ab-b 2+a2-ab+ab=0

a^2-b^2+ab=0

Divide it by ab and get a b-b a+1=0

Let b a=t

1/t-t+1=0

t^2-t-1=0

It is obtained according to the root finding formula.

t=(1±√5)/2

Because a b is negative.

So b a=(1+ 5) 2

Knowing that a and b are both negative real numbers, and 1 a+1 b+1 (a-b)=0, find the value of b a.

Solution: 1 a+1 b+1 (a-b)=0

1/a+1/b=1/(b-a）

a+b)/ab=1/(b-a）

b^2-a^2=ab

Divide both sides by ab to obtain:

b/a-a/b=1

Let b a=x, then.

x-1/x=1

x^2-x-1=0

Solution: x1=(1+ 5) 2, x2=(1- 5) 2 (undesirable) b a=(1+ 5) 2

1\a+1\b=1\(b-a)

Multiply both sides by b

b\a+1=b\(b-a)

Both sides are taken from the reciprocal.

1 (b a+1)=(b-a) b=1-a b let b a=x

Simplification yields 1 (x+1)=1-1 x

i.e. x 2 - x - 1 = 0

You can get the answer.

1\a+1\b+1\(a-b)=0

a+b)/ab+1/(a-b)=0

Shift (a+b) ab=-1 (a-b).

a^2-b^2+ab=0

both are divided by a 2

1-(b/a)~2+b/a=0

Let b a=t

1-t~2+t=0

It is obtained according to the root finding formula.

t=(1±√5)/2

Because a b is negative.

So b a=(1+ 5) 2

The inscription "1 Lingjian A+1 B-1 (A-B)=0" is sorted out from 1 A+1 B-1 (A-B)=0 A 2-b 2=ab to launch a b-b a=1 to set a b=t, then b a=1 t will be a b=t, b a=1 t into the closed Ming a b-b a=1 to get t-1 t=1, and the sedan car Wang sue to get t1=(1 + root number 5) 2, t2=(1-root number 5) 2, that is, a b = (1 + root number 5) 2 or a b = (1 -..

The inscription is "1 a+1 b-1 (a-b)=0".

From 1 a+1 b-1 (a-b)=0, a 2-b 2=ab is sorted out, and a b-b a=1 is introduced

Let a b=t, then b a=1 t

Substituting a b = t, b a = 1 t into a b a - b a = 1 gives t-1 t = 1, and the solution is t1 = (1 + root number 5) 2, t2 = (1 - root number 5) 2

That is, a b = (1 + root number 5) 2 or a b = (1 - root number 5) 2 and since a and b are both negative real numbers, a b is a positive value.

Therefore a b = (1 + root number 5) 2

1/a+1/b=1/(a-b)

a+b)/ab=1/(a-b)

a+b)(a-b)=ab

a^2-b^2=ab

a^2-ab-b^2=0

Think of a as an unknown.

Discriminant = (-b) 2 + 4b 2 = 5b 2

b<0 so (5b 2)=- 5*b

So a=[b (-5b)] 2=[(1 5) 2]*ba b=(1 5) 2

Because a<0, b<0

So a b>0

So a b = (1 + 5) 2

a+b=1, then b=1-a

m=a²+b²=a²+（1-a）²=2a²-2a+1=2(a²-a+1/4)+1/2=2(a-1/2)²+1/2≥1/2

So m has a minimum value of 1 2 and no maximum value.

∵√（a²+b²）/2）≥（a+b）/2

m=a²+b²≥1/2

There is a minimum value of 1 2 and no maximum value.

Denote two unknowns as one unknown.

3a 2b=5-c(1)

2a b=1 3c(2)

Eliminate b, (2) by 2 minus (1).

Get a = 7c-3

Similarly, b = 7-11c

Since a and b are non-negative real numbers, a 0, b 0

So 7-3c 0, 7-11c 0

C 3 7 C 7 11 3 7 C 7 11m=3(7C-3) (7-11C)-7C=3C-2 is the smallest m, so C is the smallest.

So when c = 3 7, m is the smallest.

m = 3 * (3 7) - 2 = -5 7 (minus 5/7).

1/a+1/b=1/(a-b)

b(a-b)+a(a-b)=ab

a^2-b^2-ab=0

1-(b/a)^2-b/a=0

The solution yields b a=(-1 sqrt(5)) 2

And because ab is a negative real number, so.

b/a=(-1+sqrt(5))/2

Note: sqrt denotes the root number.

Tell you a simple way to do it, and then solve it, and you're good to go.

Can the landlord write it clearly?

1\a+1\b+1\(a-b)=0

The general score is [b(a-b)+a(a-b)+ab] ab(a-b)=0, so ab-b 2+a 2-ab+ab=0

a^2-b^2+ab=0

Divide it by ab and get a b-b a+1=0

Let b a=t

1/t-t+1=0

t^2-t-1=0

It is obtained according to the root finding formula.

t=(1±√5)/2

Because a b is negative.

So b a=(1+ 5) 2

1/a+1/b-1/(a-b)=0

1/a+1/b=1/(a-b)

Multiply both sides of the equal sign by ab(a-b) to get :

b(a-b)+a(a-b)=ab

a�0�5-ab-b�0�5=0；Divide both sides of the equal sign by b 0 5(a b) 0 5-(a b)-1=0

According to the known conditions and the root finding formula, we get:

a/b=(1+√5)/2

a, b, c meet a+b+c=0, abc>0 => a, b, c two negative and one positive.

Let a<0, b<0, c>0 a+b = - c, ab < c^2

1/a+1/b+1/c

a+b) / (ab) +1/c

c /(ab) +1/c

-c^2 + ab)/(abc)

0 so B should be chosen

Because (a+b+c) = 0

Because a +b +c +2(ab+bc+ac)=0, -(a+b+c)=2(ab+bc+ac)ab+bc+ac<0

1 A+1 B+1 C=(Ab+BC+AC) ABCbecause ABC>0 AB+BC+AC<0 (AB+BC+AC) ABC<0

Mine is convincing.

(a+b+c)*(a+b+c)=a^2+b^2+c^2+2ab+2ac+2bc=0

Because abc=8

then a, b, and c are not equal to 0

So a 2 + b 2 + c 2>0

Then 2(ab+ac+bc) <0

ab+ac+bc<0

Again: 1 A+1 B+1 C=(Ab+AC+BC) abc denominator 8

Molecule 01 a+1 b+1 c<0

abc=8 two negative and one positive or three positive.

a+b+c=0 means two negative and one positive.

You can only add positive and negative to negative.

So it's negative.

The three real numbers are 4, 2, 2. The question is a bit unsmooth.