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) )