a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

`Answer:`

\(a)\left(-3x^2y-2xy^2+6\right)+\left(-x^2y+5xy^2-1\right)\)

\(=-3x^2y-2xy^2+6+-x^2y+5xy^2-1\)

\(=\left(-3x^2y-x^2y\right)+\left(-2xy^2+5xy^2\right)+\left(6-1\right)\)

\(=-4x^2y+3xy^2+5\)

\(b)\left(1,6x^3-3,8x^2y\right)+\left(-2,2x^2y-1,6x^3+0,5xy^2\right)\)

\(=1,6x^3-3,8x^2y+-2,2x^2y-1,6x^3+0,5xy^2\)

\(=\left(1,6x^3-1,6x^3\right)+\left(-3,8x^2y+-2,2x^2y\right)+0,5xy^2\)

\(=-6x^2y+0,5xy^2\)

\(c)\left(6,7xy^2-2,7xy+5y^2\right)-\left(1,3xy-3,3xy^2+5y^2\right)\)

\(=6,7xy^2-2,7xy+5y^2-1,3xy+3,3xy^2-5y^2\)

\(=\left(6,7xy^2+3,3xy^2\right)+\left(-2,7xy-1,3xy\right)+\left(5y^2-5y^2\right)\)

\(=10xy^2+-4xy\)

\(=10xy^2-4xy\)

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

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

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

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

\(e)\left(x^2+y^2-2xy\right)-\left(x^2+y^2+2xy\right)+\left(4xy-1\right)\)

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

\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(-2xy-2xy+4xy\right)-1\)

\(=-1\)

a: (x-1)(2y-4)=0

=>x-1=0 và 2y-4=0

=>x=1 và y=2

b: (3x-2)(y-3)=6

mà x,y là số nguyên

nên \(\left(3x-2;y-3\right)\in\left\{\left(1;6\right);\left(-2;-3\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;9\right);\left(0;0\right)\right\}\)

d: \(\left(3x-4\right)\left(2y-1\right)=2\)

\(\Leftrightarrow\left(3x-4;2y-1\right)\in\left\{\left(2;1\right);\left(-2;-1\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)=\left(2;1\right)\)

a) \(2xy+3z+6y+xz\)

\(=2xy+2.3y\)

\(=2y\left(x+3\right)+3z+xz\)

\(=2y\left(x+3\right)+z\left(x+3\right)\)

\(=\left(x+3\right)\left(2y+z\right)\)

c) \(x^4-9x^3+x^2-9x\)

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

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

NHƯNG THIẾU BẠN ƠI

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

c)  \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)