Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

a) 9x² - 25

= (3x)² - 5²

= (3x - 5)(3x + 5)

b) y³ + 6y² + 9y

= y(y² + 6y + 9)

= y(y² + 2.y.3 + 3²)

= y(y + 3)²

c) m² - n² - 2m + 1

= (m² - 2m + 1) - n²

= (m - 1)² - n

= (m - 1 - n)(m - 1 + n)

= (m - n - 1)(m + n - 1)

a: \(3x-3y+x^2-y^2\)

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

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

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

b: \(x^2-4x^2y^2+y^2+2xy\)

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

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

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

c: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d: \(x^3-3x^2+3x-1-y^3\)

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

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

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

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

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

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

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

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

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

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

f: \(x^3-2x^2+4x-8\)

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

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

g: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

h: \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

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

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

2: \(8xy-24xy+16x\)

\(=8x\cdot y-8x\cdot3y+8x\cdot2\)

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

\(=-16y\left(y-1\right)\)

3: \(xy-x=x\cdot y-x\cdot1=x\left(y-1\right)\)

11: \(2mx-4m2xy+6mx\)

\(=2mx-2my\cdot4y+2mx\cdot3\)

\(=2mx\left(1-4y+3\right)\)

\(=2mx\left(4-4y\right)=8mx\left(1-y\right)\)

12: \(7x^2y^5-14x^3y^4-21y^3\)

\(=7y^3\cdot x^2y^2-7y^3\cdot2x^3y-7y^3\cdot3\)

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

13: \(2\left(x-y\right)-a\left(x-y\right)\)

\(=2\cdot\left(x-y\right)-a\cdot\left(x-y\right)\)

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

a: \(-x^2+10x+3\)

\(=-\left(x^2-10x-3\right)\)

\(=-\left(x^2-10x+25-28\right)\)

\(=-\left(x^2-10x+25\right)+28\)

\(=-\left(x-5\right)^2+28< =28\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

Vậy: GTLN của \(-x^2+10x+3\) là 28 khi x=5

b: \(8x-x^2-16\)

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

\(=-\left(x-4\right)^2< =0\forall x\)

Dấu '=' xảy ra khi x-4=0

=>x=4

vậy: GTLN của \(8x-x^2-16\) là 0 khi x=4

\(x^2-10xy+25y^4\\ =x^2-2.5.x.y+\left(5y^2\right)^2\\ =\left(x-5y^2\right)^2\)

Thay \(x=105,y=5\) vào biểu thức ta được:

\(\left(105-5.5^2\right)^2\\ =\left(105-5.25\right)^2\\ =\left(-23\right)^2\\ =529\)

x^2-10xy+25y^4 = (x-5y)^2

thay x=105, y=5 ta được (105-5.5)^2=80^2=6400

\(A=3x^2+6xy+3y^2-3z^2\)

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

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

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

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

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

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

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