a) $4x^2+4x+1$

$=(2x)^2+2\cdot2x\cdot1+1^2$

$=(2x+1)^2$

b) $x^2+6x-y^2+9$

$=(x^2+6x+9)-y^2$

$=(x^2+2\cdot x\cdot3+3^2)-y^2$

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

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

$\text{#}Toru$

a: \(4x^2+4x+1\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)

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

b: \(x^2+6x-y^2+9\)

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

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

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

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

1)\(4x^2-4xy+y^2-8x+4y=\left(4x^2-4xy+y^2\right)-\left(8x-4y\right)=\left(2x-y\right)^2-4\left(2x-y\right)=\left(2x-y\right)\left(2x-y-4\right)\)

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

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

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

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

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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

\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

a: \(=x\left(x^2+4x+4-z^2\right)\)

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

\(4x^2-4x+1-\left(2x-1\right)b+\dfrac{b^2}{4}\)

\(=\left(2x-1\right)^2-2.\left(2x-1\right).\left(\dfrac{1}{2}b\right)+\left(\dfrac{b}{2}\right)^2\)

\(=\left(2x-1-\dfrac{1}{2}b\right)^2\)

Ta có: \(\left(4x^2-4x+1\right)-\left(2x-1\right)b+\dfrac{b^2}{4}\)

\(=\left(2x-1\right)^2-2\cdot\left(2x-1\right)\cdot\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2\)

\(=\left(2x-\dfrac{1}{2}b-1\right)^2\)

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

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

\(C=-25x^2+y^2-6y+9=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-\left(5x\right)^2=\left(y-3-5x\right)\left(y-3+5x\right)\)\(D=x^2-4x-y^2-8y-12=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)=\left(x-2\right)^2-\left(y+4\right)^2=\left(x-2-y-4\right)\left(x-2+y+4\right)=\left(x-y-6\right)\left(x+y+2\right)\)

a: Ta có: \(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)

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

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

b: Ta có: \(B=9x^2-\left(y^2-4y+4\right)\)

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

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

a: \(7x-14y=7\left(x-2y\right)\)

b: \(4x^2-4x+1=\left(2x-1\right)^2\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

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

e.

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

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

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2