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17 tháng 12 2017

a.5x2-10xy+5y2-20z2

  =5(x2-2xy+y2-4z2)

  =5[ (x2-2xy+y2)-(2z)]

  =5[ (x-y)2-(2z)2 ]

  =5(x-y-2z)(x-y+2z)

b.16x-5x2-3

  =15x+x-5x2-3

  =(15x-3)+(x-5x2)

  =3(5x-1)+x(1-5x)

  =3(5x-1)-x(5x-1)

  =(5x-1)(3-x)

c.x2-5x+5y-y2

  =(5y-5x)+(x2-y2)

  =5(y-x)+(x-y)(x+y)

  =5(y-x)-(y-x)(y+x)

  =(y-x)[5-(y+x)]

  =(y-x)(5-y-x)

d.3x2-6xy+3y2-12z2     (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)

=3(x2-2xy+y2-4z2)

=3[ (x2-2xy+y2)-(2z)2 ]

=3[ (x-y)2-(2z)2 ]

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

e.x2+4x+3

=x2+3x+x+3

=(x2+x)+(3x+3)

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

=(x+1)(x+3)

f.(x2+1)2-4x2

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

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

h.x2-4x-5

=x2-5x+x-5

=(x2+x)+(-5x-5)

=x(x+1)-5(x+1)

-(x+1)(x-5)

15 tháng 10 2017

a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)

b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y

c,đề bài thiếu  ẩn ở hạng tử thứ nhất ạ !

15 tháng 10 2017

b mình viết đúng rồi mà, c hạng tử 1 là x^3

13 tháng 8 2023

1)

a) => 16x2 - 8x + 1 - 8(2x2 + 3x - 4x - 6) = 15

=> 16x2 - 8x + 1 - 8(2x2 - x - 6) = 15

=> 16x2 - 8x + 1 - 16x2 + 8x + 48 = 15

=> 49 = 15 (?) (vô lí)

=> Không tìm được x thoả mãn

b) (5x - 2)(x - 2) - 4(x - 3) = x2 + 3

=> 5x2 - 10x - 2x + 4 - 4x + 12 = x2 + 3

=> 5x2 - 16x + 16 = x2 + 3

=> 4x2 - 16x + 16 = 3

=> (2x)2 - 2.2x.4 + 42 = 3

=> (2x - 4)2 = 3

=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)

Mong bạn xem lại đề bài!

13 tháng 8 2023

2) 

a) 5x2 - 10xy + 5y2 - 20z2

= 5(x2 - 2xy + y2 - 4z2)

= 5[(x - y)2 - (2z)2]

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

b) a3 - ay - a2x + xy

= a(a2 - y) - x(a2 - y)

= (a - x)(a2 - y)

c) 3x2 - 6xy + 3y2 - 12z2

= 3(x2 - 2xy + y2 - 4z2)

= 3[(x - y)2 - (2z)2]

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

d) x2 - 2xy + tx - 2ty

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

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

14 tháng 8 2017

a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3) 

b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)

\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)

c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)

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

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

e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)

f.2(x^5)-x^2-5x ( mik ko bik làm)

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

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

h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)

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

20 tháng 11 2017

\(1.5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

\(2.16x-5x^2-3\)

\(=-\left(5x^2-16x+3\right)\)

\(=-\left(5x^2-15x-x+3\right)\)

\(=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]\)

\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)

\(=-\left(x-3\right)\left(5x-1\right)\)

\(3.x^2-5x+5y-y^2\)

\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)

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

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

\(4.3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
\(5.x^2+4x+3\)

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

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

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

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

\(6.\left(x^2+1\right)^2-4x^2\)

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

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

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

\(7.x^2-4x-5\)

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

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

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

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

a) Ta có: \(3x^2+5y-3xy-5x\)

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

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

b) Ta có: \(3y^2-3z^2+3x^2+6xy\)

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

\(=3\left[\left(x+y\right)^2-z^2\right]\)

\(=3\left(x+y-z\right)\left(x+y+z\right)\)

c) Ta có: \(x^2-25-2xy+y^2\)

\(=\left(x-y\right)^2-5^2\)

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

d) Ta có: \(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

e) Ta có: \(x^2-5x+5y-y^2\)

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

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

f) Ta có: \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

a) \(39x-39y=39\left(x-y\right)\)

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

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

c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)

d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)

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

f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)

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

g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v ) 

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

h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)

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

16 tháng 9 2018

\(a,3x^2+2x=x\left(3x+2\right)\)

\(b,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

\(c,4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

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

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

16 tháng 9 2018

a ) 3x2 + 2x

= x. ( 3x + 2 )

b ) 5x - 5y + ax - ay

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

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

= ( 5 + a ) ( x - y ) 

c ) 4x2 - 25

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

d ) x2 + 6x + 5

= x2 + x + 5x + 5

= x.( x + 1 ) + 5.( x + 1 )

= ( x + 1 ) ( x + 5 )

e ) x2 - y2 + 2y - 1

= x2 - ( y - 1 )2

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

f ) x3 - 3x + 2

= x3 + 2x2 - 2x2 - 4x + x + 2

= x2 ( x + 2 ) - 2x ( x + 2 ) + ( x + 2 )

= ( x + 2 ) ( x2 - 2x + 1 )

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