Register Now

222-9+11+12:2*14+14 = ? ( )

## 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

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 – 25 + y ² – 2xy
f) 36x ² – a ² + 10a – 25
Bài 2 : Phân tích đa thức thành nhân tử:
a) x ² + x +2
b) 3x ² + 7x – 6
c) 8x ² – 23x – 3
d) -10x ² – 17x + 6
Bài 3: Tìm x, biết:
a) x ³ – x ² – 25x + 25 = 0
b) 4x ³ – 8x ² – 9x + 18 = 0
Bài 4: Tìm x
a) 3x ( x – 1 ) + x – 1 = 0
b) ( x – 2 ) ( x ² + 2x + 7 ) + 2 ( x ² – 4 ) – 5 ( x – 2 ) = 0
c) ( 2x – 1 ) ² – 25 = 0
d) x ³ + 27 + ( x + 3 ) ( x – 9 ) = 0

1. 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

2. 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