\(a,x^2+9x+20=x^2+4x+5x+20.\)

\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)

\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)

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

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

\(c,x^4+4=x^4+4x^2+4-4x^2\)

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

\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

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

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

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

a: 2x^2y-50xy=2xy(x-25)

b: 5x^2-10x=5x(x-2)

c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)

d: \(x^2-xy+x=x\left(x-y+1\right)\)

e: x(x-y)-2(y-x)

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

=(x-y)(x+2)

f: 4x^2-4xy-8y^2

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

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

=4[x(x-2y)+y(x-2y)]

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

f1: x^2ỹ-y^2+y

=(x-y)(x+y)+(x+y)

=(x+y)(x-y+1)

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

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

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

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

3) \(x^3-2x^2+1\)

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

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

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

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

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

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

5) \(x^3-6x^2+9x\) (chắc đề như vậy)

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

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

6) \(4x^3-9x^2+5x\)

\(=x\left(4x^2-9x+5\right)\)

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

\(=x\left[4x\left(x-1\right)-5\left(x-1\right)\right]\)

\(=x\left(x-1\right)\left(4x-5\right)\)

\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

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

1) \(x^4+2x^3-9x^2-10x-24\)

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

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

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

2) \(6x^4+7x^3+5x^2-x-2\)

\(=6x^4-3x^3+10x^3-5x^2+10x^2-5x+4x-2\)

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

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

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

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

3) \(2x^4+3x^3+2x^2-1\)

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

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

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

4) \(x^3-x^2-x-2\)

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

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

Ta có : x³ - 3x² - x + 3

= (x³ - 3x²) - (x - 3)

= x²(x - 3) - (x - 3)

= (x - 3)(x² - 1)

= (x - 3)(x - 1)(x + 1)

1) Ta có : x(x - 2) - x + 2 = 0

=> x(x - 2) - (x - 2) = 0

=> (x - 2)(x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

Ta có \(x^3-3x^2-x+3\)

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

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

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

g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

                                       =\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)

đặt \(x^2+7x+10=a\)

ta có \(a\left(a+2\right)-24=a^2+2a-24\)

                                      \(=a^2+2a+1-25\)

                                      \(=\left(a+1\right)^2-5^2\)

                                      \(=\left(a+1-5\right)\left(a+1+5\right)\)

                                     \(=\left(a-4\right)\left(a+6\right)\)

\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

a) = (x +5)² - 2² = (x+5 -2)(x+5 +2) = (x+3)(x+7)

b) = x(x² -1) -6(x-1)= x(x+1)(x-1) -6(x-1) = (x-1)(x(x+1)-6)

a) \(x^2+10x+21=x^2+3x+7x+21\)

                                \(=x\left(x+3\right)+7\left(x+3\right)=\left(x+7\right)\left(x+3\right)\)

từ ý b đến ý e dùng chức lăng EQN của máy tính

bạn mở mt bấm mode rồi ấn 5 rồi tìm cái có ax³+bx²+cx+d

b) nhập 1 = ; 0 =;-7=;6= rồi ấn = sao cho hết nghiệm

có \(X_1=2;X_2=1;X_3=-3\)