Ói , hoa mắt chóng mặt nhức đầu ,

sao giống có chữa quá z

a) x² + y² +2x - 4y + 5 = 0

( x² + 2x + 1 ) + ( y² - 4y + 4 ) = 0

( x + 1 )² + ( y - 2 )² = 0

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

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

\(x^2-x+\dfrac{1}{4}+4y^2-4y+1=0\)

\(\left(x-\dfrac{1}{2}\right)^2+\left(2y-1\right)^2=0\)

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

Lời giải:

Gọi biểu thức trên là $A$. Ta có:

$A=(x-2y)(x^2+2xy+4y^2)(x+2y)(x^2-2xy+4y^2)$

$=[x^3-(2y)^3][x^2+(2y)^3]$

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

$=x^6-64y^6=2^6-64.(-1)^6=64-64=0$

1/ x^2 +4xy +4y^2 = (x +2y)^2

2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3

3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3

4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3

1) x² + 4xy + 4y² = ( x + 2y )²

2) - x³ + 9x² - 27x + 27 = ( 3 - x )²

3) 8x⁶ + 36x⁴y + 54x²y² + 27y³ = ( 2x² + 3y )³

4) x³ - 6x²y + 12xy² - 8y³ = ( x - 2y )³

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

6) x² + y² + 4 + 2xy + 4x + 4y = ( x + y + 2 )²

a: \(=-xy\cdot x^2-2xy\cdot xy+3\cdot xy\)

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

b: \(=\left(5x-x+4y-4y\right)\left(y-5x\right)\)

\(=4x\left(y-5x\right)=-20x^2+4xy\)

c: \(=\left(4x^2-2y\right)\left(5x^3-4y^2\right)\)

\(=4x^2\cdot5x^3-4x^2\cdot4y^2-2y\cdot5x^3+2y\cdot4y^2\)

\(=20x^5-16x^2y^2-10x^3y+8y^3\)

M = 5 - x² + 2x - 4y² - 4y

= (- x² + 2x - 1) + (- 4y² - 4y - 1) + 7

= 7 - (x - 1)² - (2y + 1)²\(\le7\)

Dấu "=" xảy ra khi x = 1 và y = - 0,5

(^~^)

M = - x² + 2xy - 4y² + 2x + 10y - 8

- M = x² - 2xy + 4y² - 2x - 10y + 8

= (y² + 1 + x² + 2y - 2xy - 2x) + (3y^2 - 12y + 12) - 5

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

\(\Rightarrow M\le5\)

Dấu "=" xảy ra khi y = 2 và x = 3.

giúp mik vs mik k cho

mai mik kt 1 tiết r

a,

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

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

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

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

\(=0\)