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\(\hept{\begin{cases}\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\\\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\end{cases}}\)

<=> \(\hept{\begin{cases}x^3+8y^3=0\left(1\right)\\x^3-8y^3=16\left(2\right)\end{cases}}\)

Lấy (1) + (2) theo vế

=> 2x³ = 16

=> x³ = 8 = 2³

=> x = 2

Thế x = 2 vào (1)

=> 2³ + 8y³ = 0

=> 8 + 8y³ = 0

=> 8y³ = -8

=> y³ = -1 = (-1)³

=> y = -1

Vậy \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

1.\(=x^3+8y^3-x^3+8y^3+2y^3=18y^3\)

2. \(=x^3-3x^2+3x-1+1-x^3+3\left(9-x^2\right)\)

\(=-3x^2+3x+27-3x^2=3\left(x+9\right)\)

Ko chắc lém :))))

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+8y^3=0\\x^3-8y^3=16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^3=8\\y^3=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

mơn bn nhìu na!!!

uk, ko có chi. mà để cho mn tham khảo lun

\(x^3+4y^3\)

x³+8y³

Cái này áp dụng hằng đẳng thức A³+B³=(A+B)(A²-AB+B²)

Chúc bạn học tốt ^^

Sửa đề: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)

a: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)

\(=x^2-3x+4y^2-6y+4xy+19\)

\(=\left(x^2+4xy+4y^2\right)-3\left(x+2y\right)+19\)

\(=\left(x+2y\right)^2-3\left(x+2y\right)+19\)

\(=\left(-5\right)^2-3\cdot\left(-5\right)+19\)

=25+15+19=59

b: \(=x^3+x^2+8y^3+4y^2+2xy\left[3\left(x+2y\right)+2\right]+70\)

\(=x^3+8y^3+x^2+4y^2+2xy\cdot\left[3\cdot\left(-5\right)+2\right]+70\)

\(=\left(x+2y\right)^3-3\cdot x\cdot2y\left(x+2y\right)+\left(x+2y\right)^2-4xy+2xy\cdot\left(-13\right)+70\)

\(=\left(-5\right)^3+\left(-5\right)^2-6xy\cdot\left(-5\right)-4xy-26xy\)+70

\(=-125+25+70=-30\)

a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)

\(=4x^2-5y-\frac{3}{5}\)

b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)

\(=\frac{5}{2}x+\frac{17}{6}xy+3\)

c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)

\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)

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

d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)

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

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

\(=y-x\)