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29 tháng 10 2020

Bài 1: (2 điểm) Phân tích các đa thức sau thành nhân tử

a) a3 – a2c + a2b – abc

= a(a2 - ac + ab - bc)

= a[a(a - c) + b(a - c)]

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

b) (x2 + 1)2 – 4x2

= (x2 + 1)2 – (2x)2

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

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

c) x2 – 10x – 9y2 + 25

= (x2 - 10x + 25) - 9y2

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

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

d) 4x2 – 36x + 56

= 4x2 - 28x - 8x + 56

= (4x2 - 28x) - (8x - 56)

= 4x(x - 7) - 8(x - 7)

= (x - 7)(4x - 8)

= 4(x - 7)(x - 2)

22 tháng 1 2021

a) a3 - a2c + a2b - abc

= a( a2 - ac + ab - bc )

= a[ a( a - c ) + b( a - c ) ]

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

b) ( x2 + 1 )2 - 4x2

= ( x2 + 1 )2 - ( 2x )2

= ( x2 - 2x + 1 )( x2 + 2x + 1 )

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

c) x2 - 10x - 9y2 + 25

= ( x2 - 10x + 25 ) - 9y2

= ( x - 5 )2 - ( 3y )2

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

d) 4x2 - 36x + 56

= 4( x2 - 9x + 14 )

= 4( x2 - 7x - 2x + 14 )

= 4[ x( x - 7 ) - 2( x - 7 ) ]

= 4( x - 7 )( x - 2 )

22 tháng 1 2021

a,\(a^3-a^2c+a^2b-abc\)

\(=a\left(a^2-ac+ab-bc\right)\)

\(=a\left[a\left(a-c\right)+b\left(a-c\right)\right]\)

\(=a\left(a-b\right)\left(a-c\right)\)

b,\(\left(x^2+1\right)^2-4x^2\)

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

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

c,\(x^2-10x-9y^2+25\)

\(=\left(x^2-10x+25\right)-9y^2\)

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

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

d,\(4x^2-36x+56\)

\(=4\left(x^2-9x+14\right)\)

\(=4\left(x^2-7x-2x+14\right)\)

\(=4\left(x-7\right)\left(x-2\right)\)

9 tháng 7 2019

Bài 1:tìm x ,biết:

a) (2x - 1)(3x + 2) - 6x(x + 1) = 0

\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)

\(\Leftrightarrow-5x=2\)

\(\Leftrightarrow x=\frac{-2}{5}\)

b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)

\(\Leftrightarrow-10x=-4\)

\(\Leftrightarrow x=\frac{2}{5}\)

c) \(4x^2-1=2\left(2x+1\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)

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

b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)

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

c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)

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

d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)

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

e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)

\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)

\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)

2 tháng 8 2017

a)

\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)

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

b)

\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)

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

c)

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

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

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

\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)

2 tháng 8 2017

(2x-3y)(2x+3y) chớ x + 3 k ik

16 tháng 8 2018

a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)

b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)

c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)

d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)

f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)

                             \(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)

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

h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)

i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)

9 tháng 8 2017

a) 4x2-8x=0

   (2x)2-2.2.2x+4-4=0

  (2x-2)2 =4

   2x-2=2

   2x  =4

    x=2

Nhớ k cho mk nha

8 tháng 9 2019

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

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

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

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

\(=4x^2-\left(y^2-8y+16\right)\)

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

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

15 tháng 7 2018

\(x^2+4x-9y^2+4\)

\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)

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

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

\(x^2-9y^2-6y-1\)

\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)

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

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

Tham khảo nhé~

15 tháng 7 2018

a/Ta có:x2+4x-9y2+4

=x2+4x+4-(3y)2

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

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

b/Ta có:x2-9y2-6xy-1

=x2-6xy-(3y)2-1

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