a)x²-5x-14=(x²-7x)+(2x-14)=x(x-7)+2(x-7)=(x-7)(x+2)

b)4x²-3x-1=(4x²-4x)+(x-1)=4x(x-1)+(x-1)=(x-1)(4x+1)

c)x²-7xy+12y²=x²-7xy+12,25y²-0,25y²=(x-3,5y)²-0,25y²=(x-3,5y-0,5y)(x-3,5y+0,5y)=(x-4y)(x-2y)

d)x³-3x+2=(x³-x)-(2x-2)=(x-1)(x²+x)-2(x-1)=(x-1)(x²+x-2)=(x-1)(x²-2x+x-2)=(x-1)(x+1)(x-2)

a) x² - 5x - 14 = x² + 2x - 7x - 14 = x(x + 2) - 7(x + 2) = (x + 2)(x - 7)

b) 4x² - 3x - 1 = 4x² - 4x + x - 1 = 4x(x - 1) + (x - 1) = (x - 1)(4x + 1)

c) x² - 7xy + 12y² = x² - 4xy - 3xy + 12y² = x(x - 4y) - 3y(x - 4y) = (x - 4y)(x - 3y)

d) x² - 3x + 2 = x² - x - 2x - 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)

xin lỗi mik mới học lớp 7 thôi!

a) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)

b) \(4^2-3x-1=16-1-3x=15-3x=3\left(5-x\right)\)

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

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

d) y^2+y-x^2+x

= (y^2 - x^2) + (x+y)

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

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

a/ \(=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\)

b/ \(=15-3x=3\left(5-x\right)\)

c/ \(=x^2-3xy-4xy+12y^2=x\left(x-3y\right)-4y\left(x-3y\right)=\left(x-4y\right)\left(x-3y\right)\)

d/ \(=x^3-2x^2+x+2x^2-4x+2=x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)

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

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

2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)

3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)

a) x²(x-3)-4x+12

=x²(x-3)-4(x-3)

=(x-3)(x²-4)

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

b) 2a(x+y)-x-y

=2a(x+y)-(x+y)

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

c) 2x-4+5x²-10x

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

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

d) 5x²-12x-7x+14

=5x²-19x+14

e) xy-y²-3x+3y

=y(x-y)-3(x-y)

=(x-y)(y-3)

#H

Trả lời:

1, x² ( x - 3 ) - 4x + 12 = x² ( x - 3 ) - 4 ( x - 3 ) = ( x - 3 )( x² - 4 ) = ( x - 3 )( x - 2 )( x + 2 )

2, 2a ( x + y ) - x - y = 2a ( x + y ) - ( x + y ) = ( x + y )( 2a - 1 ) 

3, 2x - 4 + 5x² - 10x = ( 2x - 4 ) + ( 5x² - 10x ) = 2 ( x - 2 ) + 5x ( x - 2 ) = ( x - 2 )( 2 + 5x )

4, Sửa đề: 6x² - 12x - 7x + 14 = ( 6x² - 12x ) - ( 7x - 14 ) = 6x ( x - 2 ) - 7 ( x - 2 ) = ( x - 2 )( 6x - 7 )

5, xy - y² - 3x + 3y = ( xy - y² ) - ( 3x - 3y ) = y ( x - y ) - 3 ( x - y ) = ( x + y )( y - 3 )

Cái này chưa học bt làm mấy câu

b. x^2 + 2x - 3

= x^2 + 3x - x - 3

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

= ( x + 3 ) ( x - 1 )

\(4x^2-3x-4\)

\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)

\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)

\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)

\(x^2+2x-3\)

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

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

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

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)

đặt \(x^2+5x+5=t\)

\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)

            \(=t^2-1-24\)

            \(=t^2-25\)

            \(=\left(t-5\right)\left(t+5\right)\)

hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

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

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

học tốt

d) 

Hướng dẫn :

(x+1)(x+2)(x+3)(x+4)-24

= x(x+5)(x^2+5x+10)

P/s: có gì vào trang web mathway.com viết đa thức vào rồi nhấn " factor " là ra nhân tử nhé

bài 1:

a. x² - 5=0

=>x² = 0+5 = 5

=> x = \(\sqrt{5}\)

vậy x= \(\sqrt{5}\)

sorry biết mỗi a thôi

a) x² - 5 = 0

    x²       = 0 + 5

    x²        = 5

=> x = \(\sqrt{5}\)

Vậy ...

\(x^2-5=0\)

<=>  \(\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

<=>  \(\orbr{\begin{cases}x-\sqrt{5}=0\\x+\sqrt{5}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

a) \(x^3+4x^2-21x\)

\(=x\left(x^2+4x-21\right)\)

\(=x\left(x^2-3x+7x-21\right)\)

\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)

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

b) \(5x^3+6x^2+x\)

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

\(=x\left(5x^2+5x+x+1\right)\)

\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)

\(=x\left(x+1\right)\left(5x+1\right)\)

c) \(x^3-7x+6\)

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

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

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

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

d) \(3x^3+2x-5\)

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

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

\(=\left(x-1\right)\left(3x^2+3x+5\right)\)

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

\(=x\left(x-2\right)\left(5-3x\right)\)

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

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

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

c)\(\left(=5x^2-5xy\right)+\left(7y-7x\right)\)

\(=-5x\left(y-x\right)+7\left(y-x\right)\)

\(=\left(y-x\right)\left(7-5x\right)\)

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

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

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

e)\(=[\left(x^2\right)^2+8^2+2.x^2.8]-2.8.x^2\)

\(=\left(x^2+8\right)^2-16x^2\)

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

__Chúc bạn hc tốt ......