a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

a) \(=x^3+27-54-x^3=-27\)

b) \(=8x^3+y^3\)

A=(7-2x)(7+2x)+(2x+7)²

    =49-4x²+4x²+28x+49

   = 98+28x

B=(4x-5)²-(2x-1)(8x-5)

  = 16x²-25-((8x(2x-1))-(5(2x-1)))

  = 16x²-25-((16x²+8x)-(10x+5))

  = 16x²-25-(16x²+8x-10x-5)

  = 16x²-25-16x²-8x+10x+5

   = -20+2x

a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)

\(=7-4x^2+4x^2+28x+49\)

\(=28x+56\)

b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)

\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)

\(=16x^2-40x+25-16x^2+18x-5\)

\(=-22x+20\)

c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)

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

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

\(=\left(10x-8\right)^2\)

\(=100x^2-160x+64\)

d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)

\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)

\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)

\(=-9b^2+6bc-c^2+9b^2+c^2\)

=6bc

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

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

=1

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

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

\(1,A=\dfrac{2x+1-x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}-2\right)\\ A=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+1}\\ 2,\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\Leftrightarrow y=-x-3\)

\(A=\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}+\dfrac{y}{y-x}\left(ĐKXĐ:x\ne\pm y\right)\)

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

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

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

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

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

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

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

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)