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

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

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

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

c/ \(\left(2x^2+y\right)^2-\left(2x^2-y\right)^2\)

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

\(=4x^2\cdot2y=8x^2y\)

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

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

\(=4x^3y\cdot2y=8x^3y^2\)

a: \(\left(x+2\right)^2=x^2+2\cdot x\cdot2+2^2=x^2+4x+4\)

b: \(\left(2x+y\right)^2=\left(2x\right)^2+2\cdot2x\cdot y+y^2=4x^2+4xy+y^2\)

c: \(\left(x-3y\right)^2=x^2-2\cdot x\cdot3y+\left(3y\right)^2=x^2-6xy+9y^2\)

d: \(\left(\dfrac{1}{2}x-y\right)^2=\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot y+y^2\)

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

e: \(\left(x^2-y\right)^2=\left(x^2\right)^2-2\cdot x^2y+y^2=x^4-2x^2y+y^2\)

a) \(\left(x+2\right)^2\)

\(=x^2+2\cdot x\cdot2+2^2\)

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

b) \(\left(2x+y\right)^2\)

\(=\left(2x\right)^2+2\cdot2x\cdot y+y^2\)

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

c) \(\left(x-3y\right)^2\)

\(=x^2-2\cdot x\cdot3y+\left(3y\right)^2\)

\(=x^2-6xy+9y^2\)

d) \(\left(\dfrac{1}{2}x-y\right)^2\)

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

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

e) \(\left(x^2-y\right)^2\)

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

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

a)(x+2)²=x²+2×x×2+2²=x²+4x+2

b)(2x+y)²=(2x)²+2×2x×y+y²=4x²+4xy+y²

c)(x-3y)²=x²-2×x×3y+(3y)²=x²-6xy+9y

d)(1/2x-y)²=(1/2x)²-2×1/2x×y+y²=1/4x²-xy+y²

e)(x²-y)²=(x²)²-2×x²×y+y²=x⁴-2x²y+y²

\(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

Vậy chọn đáp án A

chọn A

a) (x-y)(2x+3y)=2x²+3xy-2xy+3y²=2x²+xy+3y²

b) (2x-1)²-(2x-1)=0

<=> 2x-1=0 <=> x=\(\dfrac{1}{2}\)

a) Ta có: (x-y)(2x+3y)

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

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

b) Ta có: \(4x^2-4x+1=2x-1\)

\(\Leftrightarrow4x^2-4x+1-2x+1=0\)

\(\Leftrightarrow4x^2-4x-2x+2=0\)

\(\Leftrightarrow4x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{1;\dfrac{1}{2}\right\}\)

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

\(4x^2+4xy+y^2\)

Đáp án là: 4x^2 + 4xy + y^2

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