a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

\(a,\left(x+3\right)^2\)

\(b,\left(x+\frac{1}{2}\right)^2\)

\(c,\left(xy^2+1\right)^2\)

Ta có : Ta có : 

3xn−2(xn+2−yn+2)+yn+2(3xn−2−yn−5)3xn−2(xn+2−yn+2)+yn+2(3xn−2−yn−5)

=3xn−2⋅xn+2−3xn−2⋅yn+2+yn+2⋅3xn−2−yn+2⋅yn−5=3xn−2⋅xn+2−3xn−2⋅yn+2+yn+2⋅3xn−2−yn+2⋅yn−5

=3x(n+2)+(n−2)+(−3xn−2⋅yn+2+yn+2⋅3xn−2)−y(n+2)+(n−5)=3x(n+2)+(n−2)+(−3xn−2⋅yn+2+yn+2⋅3xn−2)−y(n+2)+(n−5)

=3xn+2+n−2−yn+2+n−5=3xn+2+n−2−yn+2+n−5

=3x(n+n)+(2−2)−y(n+n)+(2−5)=3x(n+n)+(2−2)−y(n+n)+(2−5)

=3x2n−y2n−3

1)  \(5x^2-10x+2-x=0\)

=>   \(5x\left(x-2\right)-\left(x-2\right)=0\)

=>  \(\left(5x-1\right)\left(x-2\right)=0\)

=>  \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}}\)

2)  \(x^4-4x^3+4x^2=0\)

=>  \(x^2\left(x^2-4x+4\right)=0\)

=>  \(x^2\left(x-2\right)^2=0\)

=>  \(\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

\(=4x^2y^2-12x^3y^4+9x^2y^6\)

\(4x^2-4x+1=0\)

=>  \(\left(2x-1\right)^2=0\)

=>  \(2x-1=0\)

=>  \(2x=1\)

=>  \(x=\frac{1}{2}\)

4x² -4x +1 =0

<=> (2x)² - 2.2x.1 +1² =0

<=> (2x-1)² =0

<=> 2x-1=0

<=> 2x=1

<=> x = 1/2

Vậy x = 1/2 

a) 3x² + 6x

= 3x(x +2)

b) (x+2)² - y²

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

c) x² - 2xy + 2x + y² - 2y

= (x² - 2xy +y²) + 2(x-y)

= (x-y)² + 2(x-y)

= (x-y)(x-y+2) 

d) x² - 3x-4

= x² - 4x +  x -4

= x(x-4)+ (x-4) 

= (x-4)(x+ 1) 

x² - y² + 2yz - z²

= x² - (y² - 2yz + z²)

= x² - (y-z)²

= (x-y+z)(x+y-z) 

x² - 2xy + tx - 2ty

= x(x+t) - 2y(x+t)

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