CPhill

Simplify as

3a + 2b  = 2

3a + (2 - m)b = k

For this to have infinite solutions, the equations must be the same

This implies that  m   =  0     and   k    =  2

(xy) / (x + y)  = 1

(xz) /(x + z) =1

(yz) ( y + z)  = 3

xy = x + y  

xz = x + z   →   xz - x = z  →  x ( z - 1)  = z →  x =  z / (z -1)

yz =  3(y + z)  →   yz = 3y + 3z →  yz - 3y = 3z →  y ( z - 3)  =  3z →  y = 3z / ( z -3)

xy  = x + y

[ z / ( z -1) ] [3z / ( z -3) ] = z / (z - 1) + 3z / ( z -3)

[ 3z^2 ] / [ (z -1) ( z - 3) ] =  z (z -3) + 3z ( z -1) / [ (z -3) (z -1) ]

3z^2  = z ( z -3) +3z ( z -1)

3z^2  = z^2 - 3z + 3z^2 - 3z

z^2 - 6z  =  0

z ( z - 6 ]  = 0

z = 0   (reject)

z =  6

y = 6

x  = 6/5

I still gave you a point  for a good  effort !!!!

\(\sqrt{1-\sqrt{2+\sqrt{3+\sqrt{x}}}} \)

Note   that  even if x =  0   (as small as it  can  be),,,we have

1 - sqrt [ 2 + sqrt (3) ]   which is less than   0......so the quantity under the radical isn't real

If x >  0     then    sqrt [ 2 + sqrt (3 + sqrt (x)) ]    will still be >1  which makes the quantity under the radical still < 0] 

So.....this  can't have a domain of real numbers

Assuming "a" not equal to  0, multiply through by a

a^2/3 + a  = (a + 3) - (a^2 + 2)

(1/3)a^2  + a  = -a^2 + a + 1          simplify as

(4/3)a^2  =  1             multiply both sides by 3/4

a^2  = 3/4                  take both roots

a =  sqrt (3)  / 2      or     a =  -sqrt (3)  / 2

5N / ( N + 2)  =  6

5N  = 6(N + 2)

5N  = 6N + 12

-N =  12

N =   -12

L = H  - 5

H = L + 5

L = 2W

W = L/2

W = ( H - 5) / 2

SA = 2 [ LW + HL + HW ]

234  = 2  [  (H - 5)(H - 5)/2 + H(H  -5)  +  H (H - 5)  / 2 ]

234  = (H - 5)(H-5)  + 2H (H -5)  + H(H -5)

234 =  H^2 - 10H + 25 + 2H^2 - 10H + H^2 - 5H

4H^2  - 25H  + 25  = 234

4H^2  - 25H - 209 =  0

H  =   [25 + sqrt [ 625 -  4*4 * -209 ] ] / 8   =  [25 + sqrt [ 3969 ] / 8  = [25 + 63 ] / 8  = 88 / 8  =  11

\(\frac{(2x + 3)(x - 7)}{\sqrt{x}} \)

The numerator is  defined for  all x

The  denominator   cannnot  = 0   and  x  must be > 0

So.....the most restrictive interval  of the domain  is

 (0, inf )

Take the derivative with respect to x  and  set to  0

2x + 12  =  0

x  =  -6

Take the derivative with respect to y and set to 0

2y - 16 = 0 

y = 8

Minimum  is   (-6)^2 + (8)^2 + 12 (-6) -16(8) + 8   =   -92

Bosco is correct......no solution.....