Question

Halla el mínimo común múltiplo de cada conjunto de expresiones algebraicas.
16 xy2, 8 xy, 2 x3y
x2 + 3 x +2, x2 -4
x2 - 1, x - 1, x2 + x + 2, x2 +4 x +3
x3 + 1, x3 + x2 + x + 1, x2 + x + 1
x2 - 8 x + 7, x2 -7 x, x3 - 7 x2 + x- 7

cuando el número esta de la letra significa que es una potencia, no se como hacerlo chiquito xd

Answer 1

Respuesta:

(g) If y000 Dx2ex, t h en y00 DRx2exd x Dx2ex2Rx e xd x Dx2ex2x e xC2 e xCc1;y00 .0/ D3)2Cc1D3)c1D1, so (A) y00 D.x 22x C2/e xC1. Since R.x 22x C2/e xd x D.x 22x C2/e xR.2x 2/e xd x D.x 22x C2 /e x.2x 2/e xC2 e xD.x 24x C6/e x,(A) impli es that y0D.x 24 x C6/e xCxCc2;y0.0/ D 2)6Cc2D 2)c2D 8, so (B)y0D.x 24x C6 /e xCx8; Sin ce R.x 24x C6/e xd x D.x 24 x C6/e xR.2x 4/e xd x D.x 24 x C6/e x.2x 4/e xC2 e xD.x 26x C12/e x, (B ) i mpli es that yD.x 26x C12/e xCx228x Cc3;y .0/ D1)12 Cc3D1)c3D 11, so yD.x 26x C12/e xCx228x 11.(h) If y000 D2Csin 2x , t h en y00 D2x cos 2x2Cc1;y00 .0/ D3)  12Cc1D3)c1D72,so y00 D2x cos 2x2C72. Then y0Dx2sin 2 x4C72xCc2;y0.0/ D 6)c2D 6, soy0Dx2sin 2x4C72x6. Then yDx33Ccos 2x8C74x26x Cc3;y .0/ D1)18Cc3D1)c3D78,so yDx33Ccos 2x8C74x26x C78.(i) If y000 D2x C1, then y0 0 Dx2CxCc1;y00 .2/ D7)6Cc1D7)c1D1; so y0 0 Dx2CxC1.Then y0Dx33Cx22C.x 2/ Cc2;y0.2/ D 4)143Cc2D 4)c2D  263, so y0Dx33Cx22C.x 2/ 263. Then yDx412 Cx36C12.x 2/2263.x 2/ Cc3;y .2/ D1)83Cc3D1)c3D  53,so yDx412 Cx36C12.x 2/2263.x 2/ 53.1.2.6. (a) If yDx2.1 Cln x /, t h en y .e / De2.1 Cln e / D2 e 2;y0D2x .1 Cl n x / CxD3x C2x l n x,so y0.e / D3e C2 e l n eD5 e ; (A) y00 D3C2C2l n xD5C2l n x. Now, 3 xy 04y D3x .3x C2x ln x / 4 x 2.1 Cln x / D5x 2C2x 2l n xDx2y0 0 , from (A).(b) If yDx23Cx1, th en y .1/ D13C11D13;y0D23xC1, so y0.1/ D23C1D53; (A)y00 D23. Now x2xy 0CyC1Dx2x23xC1Cx23Cx1C1D23x2Dx2y00 , from (A).(c) If yD.1 Cx2/1=2 , th en y .0/ D.1 C02/1=2 D1;y0D x .1 Cx2/3=2 , so y0.0/ D0; (A)y00 D.2x 21/.1 Cx2/5=2 . Now, .x 21/y x .x 2C1/y 0D.x 21/.1 Cx2/1=2 x .x 2C1/.x /.1 Cx2/3=2 D.2x 21/.1 Cx2/1=2 Dy00 .1 Cx2/2from (A), so y0 0 D.x 21/y x .x 2C1/y 0.x 2C1/2.(d) If yDx21x, then y .1= 2/ D1=411= 2 D12;y0D  x .x 2 /.1 x /2, so y0.1= 2/ D.1= 2/.3= 2/.1 1=2 /2D3;(A) y00 D2.1 x /3. Now, ( B) xCyDxCx21xDx1xand (C) xy 0yD  x2.x 2/.1 x /2x21xDx2.1 x /2. From (B) and (C ), .x Cy /.xy 0y / Dx3.1 x /3Dx3

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