a) x² + 4x + 3

= x² + 3x + x +3

= ( x² + 3 ) + ( x + 3 )

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

= ( x + 3 ) ( x + 1 )

b) 4x² - 4x - 3

= 4x² + 2x - 6x - 3

= ( 4x² + 2x ) - ( 6x + 3 )

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

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

c) x² - x - 12

= x² + 3x - 4x - 12

= ( x² + 3x ) - ( 4x + 12 )

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

= ( x + 3 ) ( x - 4 )

d) 4x⁴ - 4x²y² - 8y⁴

= 4 ( x⁴ - x²y² - 2y⁴ )

Hk tốt

cảm ơn bạn

a) \(x^2+4x+3\)

= \(x^2+x+3x+3\)

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

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

b) \(4x^2-4x-3\)

= \(4x^2+2x-6x-3\)

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

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

c)\(x^2-x-12\)

= \(x^2-4x+3x-12\)

= \(x\left(x-4\right)+3\left(x-4\right)\)

= \(\left(x-4\right)\left(x+3\right)\)

câu d la 4x^2.y^2 phai ko

\(4x^4-4x^2y^2-8y^4\)

= \(4x^4-8x^2y^2+4x^2y^2-8y^4\)

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

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

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

Kết bạn với mình nha

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

b, \(4x^2+4x-3=\left(2x\right)^2+2.2x+1-4=\left(2x+1\right)^2-2^2=\left(2x+1-2\right)\left(2x+1+2\right)=\left(2x-1\right)\left(2x+3\right)\)

c, \(x^2-x-12=x^2-x+\dfrac{1}{4}-\dfrac{49}{4}=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x-\dfrac{1}{2}-\dfrac{7}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(x-4\right)\left(x+3\right)\)

d, \(4x^4+4x^2y^2-8y^4=\left(2x^2\right)^2+2.2x^2y^2+\left(y^2\right)^2-9y^4=\left(2x^2+y^2\right)^2-\left(3y^2\right)^2=\left(2x^2+y^2-3y^2\right)\left(2x^2+y^2+3y^2\right)=\left(2x^2-2y^2\right)\left(2x^2+4y^2\right)=4\left(x+y\right)\left(x-y\right)\left(x^2+2y^2\right)\)

A: Đặt P(x)=0

=>3x-5=0

hay x=5/3

b: Đặt Q(x)=0

=>-2x+6=0

hay x=3

c: Đặt M(y)=0

=>1/2y-3=0

hay y=6

d: Đặt A(x)=0

=>12-3/4x=0

=>3/4x=12

hay x=16

Bài 7

a)cho P(x) = 0

\(=>3x-5=0\Leftrightarrow3x=5\Leftrightarrow x=\dfrac{5}{3}\)

b) cho Q(x) = 0

\(=>6-2x=0\Leftrightarrow2x=6\Leftrightarrow x=3\)

c)cho M(y) = 0

\(=>\dfrac{1}{2}y-3=0\Leftrightarrow\cdot\dfrac{1}{2}y=3\Leftrightarrow y=6\)

d)cho A(x) = 0

\(=>\dfrac{-3}{4}x+12=0=>-\dfrac{3}{4}x=-12=>x=16\)

e)cho B(y) = 0

=>\(2y+15=0=>2y=-15=>y=-\dfrac{15}{2}\)

f) cho C(t) = 0

=>\(2-5t=0=>5t=2=>t=\dfrac{2}{5}\)

a) \(P\left(x\right)=3x-5\)

\(3x-5=0\)

\(\)\(3x=5\)

\(x=5:3\)

\(x=\dfrac{5}{3}\)

Vậy.......

b) \(Q\left(x\right)=6-2x\)

\(6-2x=0\)

\(2x=6\)

\(x=3\)

Vậy....

c) \(M\left(y\right)=\dfrac{1}{2}y-3\)

\(\dfrac{1}{2}y-3=0\)

\(\dfrac{1}{2}y=3\)

\(y=6\)

Vậy....

d) \(A\left(x\right)=\dfrac{-3}{4}x+12\)

\(\dfrac{-3}{4}x+12=0\)

\(\dfrac{-3}{4}x=-12\)

\(x=16\)

Vậy...

e) \(B\left(y\right)=2y+18\)

\(2y+18=0\)

\(2y=-18\)

\(y=-9\)

Vậy...

f) \(C\left(t\right)=2-5t\)

\(2-5t=0\)

\(5t=2\)

\(x=\dfrac{2}{5}\)

Vậy...

a. Đặt \(x^2-2y=a\)

ta có : \(\left(x^2-2y\right)^2-4\left(x^2-2y\right)-12=a^2-4a-12=a^2-6a+2a-12=\left(a-6\right)\left(a+2\right)\)

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

b. Đặt \(x+6=a\Rightarrow\left(x+3\right)\left(x+6\right)\left(x+9\right)+45=\left(a-3\right)a\left(a+3\right)+45\)

\(=a^3-9a+45\) nghiệm xấu quá không nhóm được ban ơi :((

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

\(A=4x^4+4x^2y^2+3x^2y^2+3y^4+4y^2\)

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

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

\(=4\left(4x^2+4y^2\right)=4\cdot4\cdot4=64\)

\(x^4+4\)

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

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

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

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

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

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

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

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

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

a) \(x^4+4=x^4+4x^2+4-4x^2\)

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

b) \(4x^4+4x^2y^2-8y^4=4x^4+4x^2y^2+y^4-9y^4\)

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

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

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

a)

\(12xy-4x^2y+8xy^2\\ =4xy\cdot\left(3-x+2y\right)\)

b)

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

c)

\(25x^2\cdot\left(y-1\right)-5x^3\cdot\left(1-y\right)\\ =-25x^2\cdot\left(1-y\right)-5x^3\cdot\left(1-y\right)\\ =\left(1-y\right)\cdot\left(-25x^2-5x^3\right)\\ =5x^2\left(1-y\right)\cdot\left(-5-x\right)\)

d)

\(3x\cdot\left(a-x\right)+4a\cdot\left(a-x\right)\\ =\left(a-x\right)\cdot\left(3x+4a\right)\)

e)

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

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

\(=\left(x+1-2y\right)\left[\left(x+1\right)^2+2y\left(x+1\right)+4y^2\right]\)

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