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

\(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow\left(2x^2+2y^2+4xy\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow2\left(x^2+y^2+2xy\right)+\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

Vì \(\left(x+y\right)^2\ge0\); \(\left(x+1\right)^2\ge0\); \(\left(y-1\right)^2\ge0\)\(\forall x,y\)

\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-y\\x=-1\\y=1\end{cases}}\)

Vậy \(x=-1\)và \(y=1\)

Bài 1:
a) (2x - y) + (2x - y) + (2x - y) + 3y
= 3(2x - y) + 3y
= 3(2x - y + 3y)
= 3(2x + 2y)
= 3.2(x + y)
= 6(x + y)

b) (x + 2y) + (x - 2y) + (8x - 3y)
= x + 2y + x - 2y + 8x - 3y
= 9x - 3y
= 3(3x - y)

c) (x + 2y) - 2(x - 2y) - (2x - 3y)
= x + 2y - 2x + 4y - 2x + 3y
= 9y - 3x
= 3(3y - x)

Bài 2:
M + 2(x² - 4y²) + Q = 6x² - 4xy + 5y² + P
M + 2x² - 8y²   -3x² + 7xy - 2y² = 6x² - 4xy + 5y² + 9x² - 6xy + 3y²
M + 2x² - 3x² - 6x² - 9x² - 8y² - 2y² - 5y² - 3y² + 7xy + 4xy + 6xy = 0
M - 16x² - 18y² + 17xy = 0
M = 16x² + 18y² - 17xy

\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)

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

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

a) Ta có: A = x² + y² - xy - 2x - 2y + 9

2A = 2x² + 2y² - 2xy - 4x - 4y + 18

2A = (x² + y² - 2xy) + (x² - 4x + 4) + (x² - 4y + 4) + 10

2A = (x - y)² + (x - 2)² + (y - 2)² + 10 \(\ge\)10 \(\forall\)x

=>A \(\ge\)5 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}\) <=> x = y = 2

Vậy MinA = 5 <=> x = y = 2

b) Ta có: 3x² + 3y² + 4xy + 2x - 2y + 2 = 0

=> (2x² + 2y² + 4xy) + (x² + 2x + 1) + (y² - 2y + 1) = 0

=> 2(x + y)² + (x + 1)² + (y - 1)² = 0

<=> \(\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}}\) 

<=> \(\hept{\begin{cases}x=-y\\x=-1\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

= x(x - 2y - 3)(x - 2y + 3)

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

3x²+3y²+4xy+2x-2y+2=0

=>2(x²+2xy+y²) + (x²+2x+1) + (y²-2y+1) = 0

=>2(x+y)²+(x+1)²+(y-1)²=0

Vì 2(x+y)²>= 0 với mọi x,y thuộc R

(x+1)² >=0 với mọi x thuộc R

(y-1)²>=0 với mọi y thuộc R

=> Dấu bằng xảy ra <=> x+y=0 ; x+1=0; y-1=0

<=> x= (-1), y=1

Vậy x=(-1) ; y=1

Học tốt nha ;)

  leftrightarrow (x+1)²+(y-1)² +2(x+y)²=0

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

Thay x=-1:y=1 vào bài là ok

làm ơn giúp e vs

\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)