a, 3x^2 + 13x + 10  

= 3x^2 + 3x + 10x + 10 

= 3x(x + 1) + 10(x + 1)

= (3x + 10)(x + 1)

b, x^2 - 10x + 21

= x^2 - 3x - 7x + 21

= x(x - 3) - 7(x - 3)

= (x - 7)(x - 3)

c, 6x^2 - 5x + 1

= 6x^2 - 3x - 2x + 1

= 3x(2x - 1) - (2x - 1)

= (3x - 1)(2x - 1)

Bạn đăng 1 lần nhiều bài như vậy làm người khác nản lắm đấy =) đơn giản bài rất dài mà mik cx ko chắc là bản thân mik có đc k hay ko nên phải nản vậy thôi :)

1a)\(3x^2+13x+10=3x^2+3x+10x+10\)

\(3x\left(x+1\right)+10\left(x+1\right)=\left(3x+10\right)\left(x+1\right)\)

b)\(x^2-10x+21=x^2-3x-7x+21\)

\(=x\left(x-3\right)-7\left(x-3\right)=\left(x-7\right)\left(x-3\right)\)

c)\(6x^2-5x+1=6x^2-3x-2x+1\)

\(=3x\left(2x-1\right)-\left(2x-1\right)=\left(3x-1\right)\left(2x-1\right)\)

1,

\(a,3x^2+13x+10=3x^2+3x+10x+10\)

\(=3x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x+10\right)\)

\(b,x^2-10x+21=x^2-7x-3x+21\)

\(=x\left(x-7\right)-3\left(x-7\right)\)

\(=\left(x-7\right)\left(x-3\right)\)

\(c,6x^2-5x+1=6x^2-2x-3x+1\)

\(=2x\left(3x-1\right)-\left(3x-1\right)\)

\(=\left(3x-1\right)\left(2x-1\right)\)

2,

\(a,x^8+4=\left(x^4\right)^2+4x^4+2^2-4x^4\)

\(=\left(x^4+2\right)^2-4x^4\)

\(=\left(x^4+2-2x^2\right)\left(x^4+2+2x^2\right)\)

\(b,x^4+18^2=\left(x^2\right)^2+36x^2+18^2-36x^2\)

\(=\left(x^2+18\right)-36x^2\)

\(=\left(x^2+18-6x\right)\left(x^2+18+6x\right)\)

\(c,x^4+3x^2+4=x^4+4x^2+4-x^2\)

\(=\left(x^2+2\right)-x^2\)

\(=\left(x^2+2-x\right)\left(x^2+2+x\right)\)

Toán Học Team 

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

a. 6x³-x²-486x+81

= 6x³-54x²+53x²-477x-9x+81

= 6x².(x-9)+53x.(x-9)-9.(x-9)

= (x-9).(6x²+53x-9)

= (x-9)(6x²+54x-x-9)

=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)

b. x³-5x²+3x+9

= x³+x²-6x²-6x+9x+9

=x².(x+1)-6x.(x+1)+9.(x+1)

=(x+1)(x²-6x+9)=(x+1)(x-3)²

c. x³+3x²+6x+4

= x³+x²+2x²+2x+4x+4

= x².(x+1)+2x.(x+1)+4.(x+1)

= (x+1)(x²+2x+4)

d. 

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

a,x² - 13x + 36

= x² - 4x - 9x + 36

= x(x - 4) - 9(x - 4)

= (x - 4)(x - 9)

b,x² + 3x - 18

= x² - 9 + 3x - 9

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

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

= (x - 3)(x + 6)

c,x² - 5x - 24

= x² + 3x - 8x - 24

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

= (x + 3)(x - 8)

d,3x² - 16x + 5

= 3x² - 15x - x + 5

= 3x(x - 5) - (x - 5)

= (x - 5)(3x - 1)

e, 8x² + 30x + 7

= 8x² + 2x + 28x + 7

= 2x(4x + 1) + 7(4x + 1)

= (4x + 1)(2x + 7)

g,2x² - 5x - 12

= 2x² - 8x + 3x - 12

= 2x(x - 4) + 3(x - 4)

= (x - 4)(2x + 3)

i,x⁴ + 4

= x⁴ + 4x² + 4 - 4x²

= (x² + 2)² - (2x)²

= (x² + 2 - 2x)(x² + 2 + 2x)

Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác

a) x^2+4x+3

b) 4x^2-4x-3

c)x^2-x-12

d) 4x^4-4x^2-8y^4

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

3)(x²+2x+4)(2-x)+x(x-3)(x+4)-x²+24=0

\(\Rightarrow\)8-x³+x(x²+4x-3x-12)-x²+24=0

\(\Rightarrow\)8-x³+x³+4x²-3x²-12x-x²+21=0

\(\Rightarrow\)-12x+29=0

\(\Rightarrow\)-12x=-29

\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)

a)\(2x^2+x-6=2x^2+4x-3x-6=\left(x+2\right)\left(2x-3\right)\)

b)\(6x^4+7x^2+2=6x^4+4x^2+3x^2+2=\left(3x^2+2\right)\left(2x^2+1\right)\)

c)\(2x^2-3x-2700=2x^2+72x-75x+2700=\left(2x-75\right)\left(x+36\right)\)

a)   \(x^2-3x+2\)

\(\Leftrightarrow x^2-2x-x+2\)

\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\)

b)  \(x^2-6x+8\)

\(\Leftrightarrow x^2-4x-2x+8\)

\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-2\right)\)

c)  \(3x^2+9x-30\)

\(\Leftrightarrow3\left(x^2+3x-10\right)\)

\(\Leftrightarrow3\left[\left(x^2+2\cdot\frac{3x}{2}+\frac{9}{4}\right)-\frac{49}{4}\right]\)

\(\Leftrightarrow3\left[\left(x+\frac{3}{2}\right)^2-\left(\frac{7}{2}\right)^2\right]\)

\(\Leftrightarrow3\left(x+\frac{3}{2}+\frac{7}{2}\right)\left(x+\frac{3}{2}-\frac{7}{2}\right)\)

\(\Leftrightarrow3\left(x-2\right)\left(x+5\right)\)

d)  \(x^2-9x+18\)

\(\Leftrightarrow x^2-3x-6x+18\)

\(\Leftrightarrow x\left(x-3\right)-6\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

TK MK NKA !!!! TH@NK !!! 

a) x² - 3x + 2 ( như này mới phân tích được ạ :) )

= x² - x - 2x + 2

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

= ( x - 2 )( x - 1 )

b) x² - 6x + 8

= x² - 2x - 4x + 8

= x( x - 2 ) - 4( x - 2 )

= ( x - 4 )( x - 2 )

c) 3x² + 9x - 30

= 3( x² + 3x - 10 )

= 3( x² - 2x + 5x - 10 )

= 3[ x( x - 2 ) + 5( x - 2 )]

= 3( x + 5 )( x - 2 )

d) x² - 9x + 18

= x² - 3x - 6x + 18

= x( x - 3 ) - 6( x - 3 )

= ( x - 6 )( x - 3 )

a, \(x^2-3x+2=x^2-x-2x+2\)

\(=x\left(x-1\right)-2\left(x-1\right)=\left(x-2\right)\left(x-1\right)\)

b, \(x^2-6x+8=x^2-2x-4x+8\)

\(=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

c, \(3x^2+9x-30=3\left(x^2+3x-10\right)\)

\(=3\left(x^2-2x+5x-10\right)=3\left(x+5\right)\left(x-2\right)\)