a,\(2x^2-8x+y^2+2y+9=0\)

\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)

\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\) 

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

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

Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)

Vậy x=2;y=-1

f) x² + 2y² - 2xy + 2x + 2 - 4y =0 
<=>x² + y² - 2xy+2x-2y+y²-2y+1+1=0 
<=>(x-y)²+2(x-y)+1+(y-1)²=0 
<=>(x-y+1)²+(y-1)²=0 
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))

2⁴ x X -3 x 5 x X = 5² - 2⁴

Bài 1

1) 4x - x² - 4 = 0

⇔ -( x² - 4x + 4 ) = 0

⇔ -( x - 2 )² = 0

⇔ x - 2 = 0

⇔ x = 2

2) 4( x - 1 )² - ( 5 - 2x )² = 0

⇔ 2²( x - 1 )² - ( 5 - 2x )² = 0

⇔ ( 2x - 2 )² - ( 5 - 2x ) = 0

⇔ ( 2x - 2 - 5 + 2x )( 2x - 2 + 5 - 2x ) = 0

⇔ ( 4x - 7 ).3 = 0

⇔ 4x - 7 = 0

⇔ x = 7/4

3) 9( x - 2 )² - 4( 3 - x )² = 0

⇔ 3²( x - 2 )² - 2²( x - 3 )² = 0

⇔ ( 3x - 6 )² - ( 2x - 6 )² = 0

⇔ ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 ) = 0

⇔ x( 5x - 12 ) = 0

⇔ x = 0 hoặc 5x - 12 = 0

⇔ x = 0 hoặc x = 12/5

4) x² - 6x + 5 = 0

⇔ x² - 5x - x + 5 = 0

⇔ x( x - 5 ) - ( x - 5 ) = 0

⇔ ( x - 5 )( x - 1 ) = 0

⇔ x - 5 = 0 hoặc x - 1 = 0

⇔ x = 5 hoặc x = 1

Bài 2.

1) x² - z² + y² - 2xy

= ( x² - 2xy + y² ) - z²

= ( x - y )² - z²

= ( x - y - z )( x - y + z )

2) a³ - ay - a²x + xy

= ( a³ - a²x ) - ( ay - xy )

= a²( a - x ) - y( a - x )

= ( a - x )( a² - y )

3) 2xy + 3z + 6y + xz

= ( 2xy + 6y ) + ( xz + 3z )

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

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

4) x² + 2xz + 2xy + 4yz

= ( x² + 2xy ) + ( 2xz + 4yz )

= x( x + 2y ) + 2z( x + 2y )

= ( x + 2y )( x + 2z )

5) ( x + y + z )³ - x³ - y³ - z³

= x³ + y³ + z³ + 3( x + y )( y + z )( x + z ) - x³ - y³ - z³

= 3( x + y )( y + z )( x + z )

1. ( 2x + y )( 4x² - 2xy + y² ) - 8x³ - y³ - 16

= [ ( 2x )³ + y³ ] - 8x³ - y³ - 16

= 8x³ + y³ - 8x³ - y³ - 16

= -16 ( đpcm )

2. ( 3x + 2y )² + ( 3x + 2y )² - 18x² - 8y² + 3

= 2( 3x + 2y )² - 18x² - 8y² + 3

= 2( 9x² + 12xy + 4y² ) - 18x² - 8y² + 3

= 18x² + 24xy + 8y² - 18x² - 8y² + 3

= 24xy + 3 ( có phụ thuộc vào biến )

3. ( -x - 3 )³ + ( x + 9 )( x² + 27 ) + 19

= -x³ - 9x² - 27x - 27 + x³ + 9x² + 27x + 243 + 19

= -27 + 243 + 19 = 235 ( đpcm )

4. ( x - 2 )³ - x( x + 1 )( x - 1 ) + 13( x - 4 )

= x³ - 6x² + 12x - 8 - x( x² - 1 ) + 13x - 52

= x³ - 6x² + 12x - 8 - x³ + x + 13x - 52

= -6x² + 26x - 60 ( có phụ thuộc vào biến )

1. (2x+y).(4x²-2xy+y²)-8x³-y³-16

=(2x)³+y³-8x³-y³-16

=-16

Vậy đa thức trên kh phụ thuộc vào biến x

2. (3x+2y)²+(3x+2y)²-18x²-8y²+3

=(9x²+12xy+4y²)+(9x²+12xy+4y²)-18x²-8y²+3

=9x²+12xy+4y²+9x²+12xy+4y²-18x²-8y²+3

=24xy+3

Vậy đa thức trên phụ thuộc biến x

ok bạn cảm ơn nha

Bài 1 : 

a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)

TH1 : \(x-3=2\Leftrightarrow x=5\)

TH2 : \(x-3=-2\Leftrightarrow x=1\)

b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

TH1 : \(x-6=0\Leftrightarrow x=6\)

TH2 : \(x+4=0\Leftrightarrow x=-4\)

c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)

d, tương tự 

Bài 2 :

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

Thay x + y = -9 ta có : 

\(\left(-9\right)^2-6\left(-9\right)-5=130\)

a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)

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

\(=x^2+y^2\)

b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)

\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)

\(=7x-9y\)

c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)

\(=\left(x+y\right)^3:\left(x+y\right)\)

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

d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)

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

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

e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)

Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)

\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)

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

f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)

\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)

\(=2x-1\)

a, (x⁴ + 2x²y² + y⁴) : (x² + y²)

= (x² + y²)² : (x² + y²)

= x² + y²

b, (49x² - 81y²) : (7x + 9y)

= (7x - 9y)(7x + 9y) : (7x + 9y)

= 7x - 9y

c, (x³ + 3x²y + 3xy² + y³) : (x + y)

= (x + y)³ : (x + y)

= (x + y)²

d, (x³ - 3x²y + 3xy² - y³) : (x² - 2xy + y²)

= (x - y)³ : (x - y)²

= x - y

Phần e thiếu thì phải

f, (8x³ - 1) : (4x² + 2x + 1)

= (2x - 1)(4x² + 2x + 1) : (4x² + 2x + 1)

= 2x - 1

Chúc bn học tốt!

Ta có : x⁴ - y⁴ 

= (x²)² - (y²)² 

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

b) 9(x - y)² - 4(x + y)²

= [3(x - y) - 4(x + y)][3(x - y) + 4(x + y)]

= [3x - 3y - 4x - 4y][3x - 3y + 4x + 4y]

= (-x - 7y)(x + y) 

e.\(x^4+2x^2+1=\left(x^2+1\right)^2\)

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

f.\(-x^2-2xy-y^2+1=-\left[\left(x+y\right)^2-1\right]=-\left(x+y-1\right)\left(x+y+1\right)=\left(x-y+1\right)\left(x+y+1\right)\)

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

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

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

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

\(=3xy\left(x+y\right)\)

tíck mình nha bn thanks !!!!!