\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)

\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)

\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)

\(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-8^2\)

\(=\left(\frac{1}{5}x+8\right)\left(\frac{1}{5}x-8\right)\)

\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3\)

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

\(8x^3+12x^2y+6xy+y^3\)

\(=2^3+3.4x^2y+3.2x.y^2+y^3\)

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

Đúng hết mà?

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

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

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

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

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

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

\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)

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

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

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

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

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

`a, 4a^2 + 4a + 1 = (2a+1)^2`

`b, -3x^2 + 6xy - 3y^2`

` = -3(x-y)^2`

`c, (x+y)^2 - 2(x+y)z + z^2`

`= (x+y-z)^2`

Bài giải:

1.

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

= -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - = (2x)³ – ()³ = (2x - )[(2x)² + 2x . + ()²]

^{= (2x - )(4x2 + x + )}

d) x² – 64y² = - (8y)² = (x + 8y)(x - 8y)

2.

a) x³ + $\frac{1}{27}$ = x³ + ($\frac{1}{3}$)³ = (x + $\frac{1}{3}$)(x² – x . $\frac{1}{3}$+ ($\frac{1}{3}$)²)

=(x + $\frac{1}{3}$)(x² – $\frac{1}{3}$x + $\frac{1}{9}$)

b) (a + b)³ – (a - b)³

= [(a + b) – (a – b)][(a + b)² + (a + b) . (a – b) + (a – b)²]

= (a + b – a + b)(a² + 2ab + b² + a² – b² + a² – 2ab + b²)

= 2b . (3a³ + b²)

c) (a + b)³ + (a – b)³ = [(a + b) + (a – b)][(a + b)² – (a + b)(a – b) + (a – b)²]

= (a + b + a – b)(a² + 2ab + b² – a² +b² + a² – 2ab + b²]

= 2a . (a² + 3b²)

d) 8x³ + 12x²y + 6xy² + y³ = (2x)³ + 3 . (2x)² . y +3 . 2x . y + y³ = (2x + y)³

e) - x³ + 9x² – 27x + 27 = 27 – 27x + 9x² – x³ = 3³ – 3 . 3² . x + 3 . 3 . x² – x³ = (3 – x)³

Bài 2 :

a) \(\left(5x^2y-8xy^2+y^3\right)\left(2x^3+x^2y-3y^2\right)\)

\(=10x^5y+5x^4y^2-15x^2y^3-16x^4y^2-8x^3y^3+24xy^4+2x^3y^3+x^2y^4-3y^5\)

\(=10x^5y-11x^4y^2-6x^3y^3+x^2y^4-15x^2y^3+24xy^4-3y^5\)

tách nhỏ câu hỏi ra bạn

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

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

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)