Ta có B = x² - 2xy + 2y² + 2y - 1 

= x² - 2xy + y² + y² + 2y + 1 - 2

= (x - y)² + (y + 1)² - 2 \(\ge-2\)

=> Min B = -2

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

Vậy Min B = -2 <=> x = y = -1

c) Ta có C = x² - 4xy + 5y² - 22y + 10x + 28

= x² - 4xy + 4y² + 10x - 20y + 25 + y² - 2y + 1 + 2 

= (x - 2y)² + 10(x - 2y) + 25 + (y - 1)² + 2 

= (x - 2y + 5)² + (y - 1)² + 2 \(\ge2\)

=> Min C = 2

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min C = 2 <=> x = -3 ; y = 1

B = ( x² - 2xy + y² ) + ( y² + 2y + 1 ) - 2 = ( x - y )² + ( y + 1 )² - 2 ≥ -2 ∀ x,y 

Dấu "=" xảy ra <=> x = y = -1 . Vậy MinB = -2

C = ( x² - 4xy + 4y² + 10x - 20y + 25 ) + ( y² - 2y + 1 ) + 2 = ( x - 2y + 5 )² + ( y - 1 )² + 2 ≥ 2 ∀ x,y

Dấu "=" xảy ra <=> x = -3 ; y = 1 . Vậy MinC = 2

Trả lời:

1, 5 ( x + y )² + 15 ( x + y ) 

= 5( x + y ) ( x + y + 3 )

2, ( a - b )² - ( a + b )( b - a )

= ( a - b )² + ( a + b )( a - b )

= ( a - b )( a - b + a + b )

= 2a ( a - b )

\(A=9y^2+12x^2-6xy+12y-6x+200\)

Rút gọn A ta có:

\(A=y\left(9y-12\right)+x\left(12x-6-6y\right)+200\)

\(MinA=0\Leftrightarrow y\left(9y-12\right)+x\left(12x-6-6y\right)=-200\)

Trả lời:

7, 5( x + y )² + 15( x + y )

= 5( x + y )( x + y + 3 )

9, 7x( y - 4 )² - ( 4 - y )³ 

= 7x ( 4 - y )² - ( 4 - y )

= ( 4 - y )² ( 7x - 4 + y )

11, ( x + 1 )( y - 2 ) - ( 2 - y )²

= ( x + 1 )( y - 2 ) - ( y - 2 )²

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

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

8, 9x ( x - y ) - 10 ( y - x )² 

= 9x ( x - y ) - 10 ( x - y )²

= ( x - y )[ ( 9x - 10 ( x - y ) ]

= ( x - y )( 9x - 10x + 10y )

= ( x - y )( 10y - x )

10, ( a - b )² - ( a + b )( b - a ) 

= ( b - a )² - ( a + b )( b - a )

= ( b - a )( b - a - a - b )

= - 2a( b - a )

= 2a ( a - b )

12, 2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )

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

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

a. \(-4x^2-4x-2\)

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

\(=-\left[\left(2x\right)^2+2.2x+1+1\right]\)

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

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

vì \(\left(2x+1\right)^2\ge0\forall x\Rightarrow-\left(2x+1\right)^2\le0\forall x\)

\(\Rightarrow-\left(2x+1\right)^2-1\le-1\forall x\)hay \(-4x^2-4x-2< 0\)

b. \(x^2+x+1\)

\(=x^2+2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

chúc bạn học tốt, k đúng cho mình nha, cảm ơn b

a)

\(x^2+6x+9\)

\(\Leftrightarrow\left(x+3\right)^2\)

b) \(\frac{1}{4}x^2-\frac{1}{2}x+1\)

\(\Leftrightarrow\left(\frac{1}{2}-x\right)^2\)

#Cừu

\(\left(ab+cd\right)^2=\left(ab\right)^2+\left(cd\right)^2=a^2b^2c^2d^2=ab.ab+cd.cd\ge k\)abcd

\(\Leftrightarrow Max\)\(k=15\)

phân tích đa thức thành nhân tử hả bạn ?

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

b, \(x^2-4x+4-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)

đề bài là gì vậy bạn