a: Ta có: \(A=x^2-20x+5\)

\(=\left(x-10\right)^2-95\ge-95\forall x\)

Dấu '=' xảy ra khi x=10

b: Ta có: \(A=x^2-5x-12\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{73}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{73}{4}\ge-\dfrac{73}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

i. Ta có: (x+y)^3 - (x-y)^3=[(x+y)-(x-y)][(x+y)^2 + (x+y)(x-y) + (x-y)^2]

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

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

có hằng đẳng thức (x+y)^3 vs (x-y)^3 thì sao bạn phải làm như vậy nữa

bạn xem mình làm như này có đúng không:

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

= 2 . 3x^2y = 6x^2y

\(x^3-x^2-5x+125\)

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

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

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^6-x^4-9x^3+9x^2\)

\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)

\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)

\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)

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

\(x^4-4x^3+8x^2-16x+16\)

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

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

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

\(3a^2-6ab+3b^2-12c^2\)

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

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

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

a) \(x^3-x^2-5x+125=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)

b) \(x^6-x^4-9x^3+9x^2=x^2\left(x^4-x^2-9x+9\right)=x^2\left[x^2\left(x-1\right)\left(x+1\right)-9\left(x-1\right)\right]=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)

c) \(x^4-4x^3+8x^2-16x+16=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x^3-2x^2+4x+8\right)=\left(x-2\right)\left(x^2+4\right)\left(x-2\right)=\left(x-2\right)^2\left(x^2+4\right)\)

d) \(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left(a-b-2c\right)\left(a-b+2c\right)\)

tick cho mk vs nha

a) 2x^2 -  2xy - 5x +5y

= (2x^2 - 2xy ) - ( 5x- 5 y)

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

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

b)8x²  +4xy-2ax-ay

=(8x² +4xy) -(2ax+ay)

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

=(2x+y).(4x-a)

c)=x(x² -4x +4)

=x(x-2)²

d)=16- (x² -2xy +y^2)

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

=(4-x+y).(4+x-y)

các câu còn lại tg tự

chúc bn hok tốt

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(A=x^5-5x^4+5x^3-5x^2+5x-6\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x-2\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-2\)

\(=-2\)

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

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

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

a)   \(x^3+3x^2+6x+4\)

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

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

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

b) \(3a^2c^2+bd+3abc+acd\)

\(=\left(3a^2c^2+acd\right)+\left(3abc+bd\right)\)

\(=ac\left(3ac+d\right)+b\left(3ac+d\right)\)

\(=\left(ac+b\right)\left(d+3ac\right)\)

c)  \(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)\)

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

d) \(x^2+y^2-x^2y^2+xy-x-y\)

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

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

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

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

e)  \(a^6-b^6=\left(a^3-b^3\right)\left(a^3+b^3\right)=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)