Toán Lớp 8: Bài 1: Phân tích đa thức thành nhân tử: a) 5x ² – 5xy – 7x + 7y b) ab – ac – b ² + bc c) 3x ³ – 12x d) x ² – 4 + ( x + 2 ) ² e) x ² – 2

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1. a)5x ² – 5xy – 7x + 7y

   =5x(x-y) – 7(x-y)

   =(x-y)(5x-7)

   b) ab – ac – b ² + bc

   =a(b-c) – b(b-c)

   =(a-b)(b-c)

   c)3x³-12x

   =3x(x²-4)

   =3x(x-2)(x+2)

   d)x² – 4 + (x+2)²

    =(x-2)(x+2) + (x+2)²

   =(x+2)(x-2+x+2)

   =2x(x+2)

   e) x ² – 25 + y ² – 2xy

   =(x² -2xy + y²)-25

   =(x-y)² – 5²

   =(x-y-5)(x-y+5)

   f)36x ² – a ² + 10a – 25

   =36x ² – (a ² – 10a + 25)

   =36x² – (a-5)²

   =[6x – (a-5)][6x+(a-5)]

   =[6x-a+5][6x+a -5]

   Bài 2 : Phân tích đa thức thành nhân tử:

   a) x ² + x -2

   =x² -x + 2x -2

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

   =(x-1)(x+2)

   b) 3x ² + 7x – 6

   =3x² +9x -2x -6

   =3x(x+3)-2(x+3)

   =(x+3)(3x-2)

   c) 8x ² – 23x – 3

   =8x² -24x + x-3

   =8x(x-3)+(x-3)

   =(x-3)(8x+1)

   d)-10x ² – 17x + 6

   =-10x² -20x +3x +6

   =-10x(x+2)+3(x+2)

   =(x+2)(-10x+3)

   Bài 3: Tìm x, biết:

   a) x ³ – x ² – 25x + 25 = 0

   ⇔x²(x-1)-25(x-1)=0

   ⇔(x-1)(x²-25)=0

   ⇔(x-1)(x-5)(x+5)=0

   ⇔x=1 hoặc x=5 hoặc x=-5

   b) 4x ³ – 8x ² – 9x + 18 = 0

   ⇔ 4x²(x-2)-9(x-2)=0

   ⇔ (x-2)(4x²-9)=0

   ⇔ (x-2)(2x-3)(2x+3)=0

   ⇔ x=2 hoặc x=$\frac{3}{2}$ hoặc x=$\frac{-3}{2}$

   Bài 4: Tìm x 

   a) 3x ( x – 1 ) + x – 1 = 0

   ⇔ (x-1)(3x+1)=0

   ⇔ x=1 hoặc x=$\frac{-1}{3}$

   b) ( x – 2 ) ( x ² + 2x + 7 ) + 2 ( x ² – 4 ) – 5 ( x – 2 ) = 0

   ⇔(x – 2)(x ² + 2x + 7) + 2(x -2)(x+2) – 5(x – 2)=0

   ⇔ (x-2)[x ² + 2x + 7+2x-4-5]=0

   ⇔ (x-2)[x²+4x-2]=0

   ⇔ x-2=0 hoặc x²+4x-2=0

   Mà x²+4x-2≠ 0 với mọi x

   ⇒ x=2

   c)(2x-1)²-25=0

   ⇔ (2x-1-5)(2x-1+5)=0

   ⇔ 2x-1= 5 hoặc 2x-1=-5

   ⇔x=3 hoặc x=-2

   Vậy x=3 hoặc x=-2

   d) x ³ + 27 + ( x + 3 ) ( x – 9 )=0

   ⇔ (x+3)(x²-3x+9)+(x+3)(x-9)=0

   ⇔ (x+3)(x²-3x+9+x-9)=0

   ⇔ (x+3)(x²-2x)=0

   ⇔ (x+3)x(x-2)=0

   ⇔ x=-3 hoặc x=0 hoặc x=2

   Trả lời
2. Bài 1:

   a)5x²-5xy-7x+7y

   =5x(x-y)-7(x-y)

   =(x-y)(5x-7)

   b)ab-ac-b²+bc

   =a(b-c)-b(b-c)

   =(b-c)(a-b)

   c)3x³-12x=3x(x²-4)=3x(x+2)(x-2)

   d)x²-4+(x+2)²

   =(x+2)(x-2)+(x+2)²

   =(x+2)(x-2+x+2)

   =2x(x+2)

   e)x²-25+y²-2xy

   =(x²-2xy+y²)-25

   =(x-y)²-5²

   =(x-y+5)(x-y-5)

   f)36x²-a²+10a-25

   =36x²-(a²-10a+25)

   =(6x)²-(a-5)²

   =(6x+a-5)(6x-a+5)

   Bài 2:

   a)x²+x-2

   =x²+2x-x-2

   =x(x+2)-(x+2)

   =(x+2)(x-1)

   b)3x²+7x-6

   =3x²+9x-2x-6

   =3x(x+3)-2(x+3)

   =(x+3)(3x-2)

   c)8x²-23x-3

   =8x²-24x+x-3

   =8x(x-3)+(x-3)

   =(x-3)(8x+1)

   d)-10x²-17x+6

   =-10x²-20x+3x+6

   =-10x(x+2)+3(x+2)

   =(x+2)(-10x+3)

   Bài 3:

   a)x³-x²-25x+25=0

   ⇔x²(x-1)-25(x-1)=0

   ⇔(x-1)(x²-25)=0

   ⇔(x-1)(x+5)(x-5)=0

   ⇔\(\left[ \begin{array}{l}x-1=0\\x+5=0\\x-5=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=1\\x=-5\\x=5\end{array} \right.\)

   Vậy x=1 hoặc x=-5 hoặc x=5

   b)4x³-8x²-9x+18=0

   ⇔4x²(x-2)-9(x-2)=0

   ⇔(x-2)(4x²-9)=0

   ⇔(x-2)(2x+3)(2x-3)=0

   ⇔\(\left[ \begin{array}{l}x-2=0\\2x+3=0\\2x-3=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x-2=0\\x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{array} \right.\)

   Vậy x=0 hoặc x=-3/2 hoặc x=3/2

   Bài 4:

   a)3x(x-1)+x-1=0

   ⇔3x(x-1)+(x-1)=0

   ⇔(x-1)(3x+1)=0

   ⇔\(\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{3}\end{array} \right.\)

   vậy x=1 hoặc x=-1/3

   b)(x-2)(x²+2x+7)+2(x²-4)-5(x-2)=0

   ⇔(x-2)(x²+2x+7)+2(x+2)(x-2)-5(x-2)=0

   ⇔(x-2)[x²+2x+7+2(x+2)-5]=0

   ⇔(x-2)(x²+2x+7+2x+4-5)=0

   ⇔(x-2)(x²+4x+6)=0

   ⇔\(\left[ \begin{array}{l}x-2=0\\x²+4x+6=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=2\\x²+4x+6=0( vô nghiệm)\end{array} \right.\)

   Vậy x=2

   c)(2x-1)²-25=0

   ⇔(2x-1)²-5²=0

   ⇔(2x-1+5)(2x-1-5)=0

   ⇔(2x+4)(2x-6)=0

   ⇔\(\left[ \begin{array}{l}2x+4=0\\2x-6=0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}2x=-4\\2x=6\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)

   Vậy x=-2 hoặc x=3

   d)x³+27+(x+3)(x-9)=0

   ⇔(x+3)(x²-3x+9)+(x+3)(x-9)=0

   ⇔(x+3)(x²-3x+9+x-9)=0

   ⇔(x+3)(x²-2x)=0

   ⇔x(x+3)(x-2)=0

   (1)x=0

   (2)x+3=0⇔x=-3

   (3)x-2=0⇔x=2

   vậy x=0 hoặc x=-3 hoặc x=2

   Trả lời