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

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

`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.

`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.

`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.

\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a. x² + 6x + 9 = (x + 3)²

b. 25 + 10x + x² = (5 + x)²

c. x² + 8x + 16 = (x + 4)²

d. x² + 14x + 49 = (x + 7)²

e. 4x² + 12x + 9 = (2x + 3)²

f. 9x² + 12x + 4 = (3x + 2)²

h. 16x² + 8 + 1 = (4x + 1)²

i. 4x² + 12xy + 9y² = (2x + 3y)²

k. 25x² + 20xy + 4y² = (5x + 2y)²

a) \(=\left(x+3\right)^2\)

b) \(=\left(x+5\right)^2\)

c) \(=\left(x+4\right)^2\)

d) \(=\left(x+7\right)^2\)

e) \(=\left(2x+3\right)^2\)

f) \(=\left(3x+2\right)^2\)

h) \(=\left(4x+1\right)^2\)

i) \(=\left(2x+3y\right)^2\)

k) \(=\left(5x+2y\right)^2\)

\(=-16x^2+24xy+6xy-9y^2\)

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

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

k

\(30xy-16x^2-9y^2\)

\(=-\left(16x^2-24xy+9y^2\right)+6xy\)

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

\(=-\left[\left(4x-3y\right)^2-6xy\right]\)

\(=-\left(4x-3y-\sqrt{6xy}\right)\left(4x-4y+\sqrt{6xy}\right)\)

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(x^2+8x+16=\left(x+4\right)^2\)

c) \(x^2+6x+9=\left(x+3\right)^2\)

d) \(4x^2+4x+1=\left(2x+1\right)^2\)

e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)

f) \(4x^2+12x+9=\left(2x+3\right)^2\)

g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)

h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)

a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)²

b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)²

c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)²

d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)²

\(1.\)

\(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^3z-x^2z^2+x^2yz-xyz^2\)

\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)

\(=\left(x^2z-xyz\right)\left(x-z\right)\)

\(=xz\left(x-y\right)\left(x-z\right)\)

\(2.\)

\(x^2-\left(a+b\right)xy+aby^2\)

\(=x^2-axy-bxy+aby^2\)

\(=x^2-bxy-axy+aby^2\)

\(=x\left(x-by\right)-ay\left(x-by\right)\)

\(=\left(x-ay\right)\left(x-by\right)\)

\(3.\)

\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)

\(=abx^2+aby^2+a^2xy+b^2xy\)

\(=abx^2+b^2xy+a^2xy+aby^2\)

\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)

\(=\left(ax+by\right)\left(bx+ay\right)\)

\(4.\)

\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)

\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)

\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)

\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)

\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)

\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)

\(5.\)

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

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

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

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

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

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

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

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

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

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

\(6.\)

\(16x^2-40xy+2y^2\)

\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)

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

\(7.\)

\(25x^4-10x^2y+y^2\)

\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)

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

\(8.\)

\(-16x^4y^6-24x^5y^5-9x^6y^4\)

\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)

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

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

\(9.\)

\(16x^2-4y^2-8x+1\)

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

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

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

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

\(10.\)

\(49x^2-25+42xy+9y^2\)

\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)

\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)

\(=\left(7x+3y\right)^2-5^2\)

\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)

2) (a-1)²+(b-2)²+(2c-1)²=0

do (a-1)², (b-2)² và (2c-1)² lớn hơn hoặc bằng 0 nên để thỏa mãn biểu thức trên thì (a-1)², (b-2)² và (2c-1)² đồng thời bằng 0

suy ra a=1, b=2, c=1/2