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

b) => 5x(x - 1) - (x - 1) = 0
    => (5x - 1) (x - 1) = 0
     5x - 1 =0  => x = 1/5 
hoặc x - 1 = 0  => x = 1
  
c) => x (x - 1) - 121 =0
    => (x - 121) (x - 1)  = 0
    x - 121 = 0  => x = 121
hoặc x - 1 = 0  => x = 1
         
   

x² - 3x + 2
= x² - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)

3x² - 7x - 10
= 3x² + 3x - 10x - 10
= 3x(x + 1) - 10(x + 1)
= (x + 1)(3x - 10)

2x² - 5x - 7
= 2x² + 2x - 7x - 7
= 2x(x + 1) - 7(x + 1)
= (x + 1)(2x - 7)

1. 2xy² +x²y⁴+1 = (xy²+1)²

2. a)3x²+3x-10x-10=3x(x+1)-10(x+1)=(x+1)(3x-10)

b)2x²-5x-7=2x²+2x-7x-7=2x(x+1)-7(x+1)=(x+1)(2x-7)

Mong có thể giúp được bạn

a,\(-4x^2+4x-1\)

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

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

\(\Rightarrow\left[2x+1-2\left(x-1\right)\right].\left[2x+1+2\left(x-1\right)\right]\)

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

\(\Rightarrow3\left(4x-1\right)\)

c,\(\left(2x-y\right)^2-4x^2+12x-9\)

\(\Leftrightarrow\left(2x+y\right)^2-\left(4x^2-12x+9\right)\)

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

\(\Leftrightarrow\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)

\(\Rightarrow\left(y+3\right)\left(4x+y-3\right)\)

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

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

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

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

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

\(25x^2+16y^2=50xy\)

\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)

\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)

Mặt khác, ta cũng có:  \(25x^2+16y^2=50xy\)

\(\Leftrightarrow\)  \(\left(5x-4y\right)^2=10xy\)

Do đó:

\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)

Vậy,  \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)

1)

 \(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)

\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)

\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)

a: Sửa đề: x^3-x^2+5x-5

=x^2(x-1)+5(x-1)

=(x-1)(x^2+5)

b: x^3+4x^2+x-6

=x^3-x^2+5x^2-5x+6x-6

=(x-1)(x^2+5x+6)

=(x-1)(x+2)(x+3)

c: \(=\left(x+2\right)^3+y^3\)

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

2.a là x=0 , x=-1, x=-2
2.b là x=2/3 , x=-5

a)x⁴+2x³+x²=x²(x²+2x+1)=x².(x+1)²

b)x²+5x-6 =x²-x+6x-6 =x.(x-1)+6.(x-1) =(x-1).(x+6)