\(A=\left(\dfrac{1}{2}x-y\right)^2=\dfrac{1}{4}x^2-xy+y^2\)

Vậy chọn B

Bài 1 :

Tự phân tích vế trái và điền vào vế phải 

Bài 2 :

a) \(3x^3-6x^2+3x\)

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

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

b) \(2xy+z+2x+yz\)

\(=\left(2xy+2x\right)+\left(z+yz\right)\)

\(=2x\left(y+1\right)+z\left(y+1\right)\)

\(=\left(y+1\right)\left(2x+z\right)\)

c) \(x^4-y^4\)

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

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

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

d) \(3x^2-4x-7\)

\(=3x^2+3x-7x-7\)

\(=3x\left(x+1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-7\right)\)

B

B

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

\(=4x^4-4x^2\)

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

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

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

d. \(\left(2x^{3y}+y\right)^2-\left(y-2x^{3y}\right)^2\)

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

\(=2y.2.2x^{3y}=4y.2x^{3y}\)

bạn ghi rõ đề ra được không

a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)

\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)

\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)

b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)

\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)

\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)

c: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)

\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)

\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)