a) (dfrac1x - 1 - dfrac3x^2x^3 - 1 = dfrac2xx^2 + x + 1)

b) (dfrac3left( x - 1 ight)left( x - 2 ight) + dfrac2left( x - 3 ight)left( x - 1 ight) )(,= dfrac1left( x - 2 ight)left( x - 3 ight))

c) (1 + dfrac1x + 2 = dfrac128 + x^3)

d) (dfrac13left( x - 3 ight)left( 2x + 7 ight) + dfrac12x + 7)(, = dfrac6left( x - 3 ight)left( x + 3 ight))

### Lời giải

a) (dfrac1x - 1 - dfrac3x^2x^3 - 1 = dfrac2xx^2 + x + 1)

Ta có: (x^3 - 1 = left( x - 1 ight)left( x^2 + x + 1 ight))

( = left( x - 1 ight)left( x^2 + x + dfrac14 + dfrac34 ight))

( = left( x - 1 ight)left< x^2 + 2.x.dfrac12 + left( dfrac12 ight)^2 + dfrac34 ight>)

(= left( x - 1 ight)left< left( x + dfrac12 ight)^2 + dfrac34 ight>)

Ta có:(left( x + dfrac12 ight)^2 geqslant 0) với mọi(x inmathbb R) đề nghị (left( x + dfrac12 ight)^2 + dfrac34 > 0)với mọi(x inmathbb R)

Do đó: (x^3 - 1 e 0)khi (x - 1 ≠ 0⇔ x ≠ 1)

ĐKXĐ: (x ≠ 1)

MTC:(x^3 - 1)

( Leftrightarrow dfracx^2 + x + 1x^3 - 1 - dfrac3x^2x^3 - 1 = dfrac2xleft( x - 1 ight)x^3 - 1)

(Rightarrow x^2 + x + 1 - 3x^2 = 2xleft( x - 1 ight) )

(Leftrightarrow - 2x^2 + x + 1 = 2x^2 - 2x)

( Leftrightarrow 0 = 2x^2 - 2x + 2x^2 - x - 1)

( Leftrightarrow 0 = 4x^2 - 3x - 1)

(Leftrightarrow 4x^2 - 3x - 1 = 0)

(Leftrightarrow 4x^2 - 4x+x - 1 = 0)

(Leftrightarrow 4xleft( x - 1 ight) + left( x - 1 ight) = 0)

(Leftrightarrow left( x - 1 ight)left( 4x + 1 ight) = 0)

( Leftrightarrow left< egingathered x - 1 = 0 hfill \ 4x + 1 = 0 hfill \ endgathered ight.)

( Leftrightarrow left< egingathered x = 1 hfill \ 4x = - 1 hfill \ endgathered ight.)

(Leftrightarrow left< matrixx = 1 ext( loại) cr x = - dfrac14 ext(thỏa mãn)cr ight.)

Vậy phương trình gồm nghiệm duy nhất (x = - dfrac14)

b) (dfrac3left( x - 1 ight)left( x - 2 ight) + dfrac2left( x - 3 ight)left( x - 1 ight) )(,= dfrac1left( x - 2 ight)left( x - 3 ight))

ĐKXĐ: (x ≠ 1, x ≠ 2, x ≠ 3)

MTC: ((x-1)(x-2)(x-3))

(Rightarrow 3left( x - 3 ight) + 2left( x - 2 ight) = x - 1)

(Leftrightarrow 3x - 9 + 2x - 4 = x - 1)

( Leftrightarrow 5x - 13 = x - 1)

( Leftrightarrow 5x - x = - 1 + 13)

(⇔ 4x = 12)

( Leftrightarrow x = 12:4)

(⇔ x = 3) (không thỏa mãn nhu cầu ĐKXĐ)

Vậy phương trình vô nghiệm.

c) (1 + dfrac1x + 2 = dfrac128 + x^3)

Ta có: (8 + x^3 =2^3 + x^3)(,=left( x + 2 ight)left( x^2 - 2x + 4 ight))

( = left( x + 2 ight)left( x^2 - 2x + 1 + 3 ight))

( = left( x + 2 ight)left< left( x - 1 ight)^2 + 3 ight>)

(left( x - 1 ight)^2 geqslant 0) với đa số (xin mathbb R) nên(left( x - 1 ight)^2 + 3 > 0)với những (xin mathbb R)

Do đó: (8 + x^3 e 0)khi (x + 2 ≠ 0 ⇔ x ≠ -2)

ĐKXĐ: (x ≠ -2)

MTC:(8 + x^3)

(Leftrightarrow dfrac8 + x^38 + x^3 + dfracx^2 - 2x + 48 + x^3 = dfrac128 + x^3)

(Rightarrow x^3 + 8 + x^2 - 2x + 4 = 12 )

( Leftrightarrow x^3 + x^2 - 2x = 12 - 8 - 4)

(Leftrightarrow x^3 + x^2 - 2x = 0)

(Leftrightarrow xleft( x^2 + x - 2 ight) = 0)

(Leftrightarrow xleft< x^2 + 2x - x - 2 ight> = 0)

⇔(x< x(x+2) - (x+2) > = 0)

⇔ (x(x + 2)(x - 1) = 0)

( Leftrightarrow left< eginarraylx = 0\x + 2 = 0\x - 1 = 0endarray ight. )

(Leftrightarrow left< eginarraylx = 0left( ext thỏa mãn ight)\x = - 2left( ext loại ight)\x = 1left( ext thỏa mãn ight)endarray ight.)

Vậy phương trình bao gồm tập nghiệm là (S = left 0;1 ight\).

d) (dfrac13left( x - 3 ight)left( 2x + 7 ight) + dfrac12x + 7 )(,= dfrac6left( x - 3 ight)left( x + 3 ight))

ĐKXĐ: (x e 3,x e - 3,x e - dfrac72)

MTC: (left( x - 3 ight)left( x + 3 ight)left( 2x + 7 ight))

(Rightarrow 13left( x + 3 ight) + left( x - 3 ight)left( x + 3 ight) )(= 6left( 2x + 7 ight) )