a) \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xz\right)\)

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

b) \(x^2-6x+9-9y^2\)

\(=\left(x^2-2\cdot x\cdot3+3^2\right)-\left(3y\right)^2\)

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

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

c) \(x^2+9x+20\)

\(=x^2+5x+4x+20\)

\(=x\left(x+5\right)+4\left(x+5\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d) \(x^4+4\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2+4-2\cdot x^2\cdot2\)

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

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

a/\(x^3+x^2y-x^2z-xyz\)

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

\(=x^2\left(x-z\right)+xy\left(x-z\right)\)

\(=\left(x-z\right)\left(x^2+xy\right)\)

b/\(x^2-6x+9-9y^2\)

\(=\left(x^2-6x+9\right)-9y^2\)

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

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

c/\(x^2+9x+20\)

\(=x^2+4x+5x+20\)

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

\(=x\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d/\(x^4+4\)

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

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

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

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

1) \(=\left(9x^2-25y^2\right)-\left(6x-10y\right)=\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)=\left(3x-5y\right)\left(3x+5y-2\right)\)

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

mk thấy A rồi nhưng chưa thấy Min

cho mk tham khảo đc ko

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

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

\(=2\left[\left(x-y\right)^2+2\left(x-y\right)\frac{1}{2}+\frac{1}{4}\right]+\frac{7}{2}\)

\(=2\left(x-y+\frac{1}{2}\right)^2+\frac{7}{2}\)

\(\Rightarrow A\ge\frac{7}{2}\)

Dấu = bn tự tính nhé

Bài làm

a) xy + y² - x - y

= ( xy + y² ) - ( x + y )

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

= ( x + y )( y - 1 )

b) 25 - x² + 4xy - 4y²

= 25 - ( x² - 4xy + 4y² )

= 25 - ( x - 2y )²

= ( 5 - x + 2y )( 5 + x - 2y )

c) xy + xz - 2y - 2z

= ( xy + xz ) - ( 2y + 2z )

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

= ( y + z )( x - 2 )

d) x² - 6xy + 9y² - 25z²

= ( x² - 6xy + 9y² ) - 25z²

= ( x - 3y )² - 25z²

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

e) 3x² - 3y² - 12x + 12y

= 3( x - y )( x + y ) - 12( x - y )

= ( x - y )[ 3( x + y ) - 12 ]

f) 4x³ + 4xy² + 8x²y - 16x

= 4x( x² + y² + 2xy - 4 )

= 4x[ ( x + y)² - 4 ]

= 4x( x + y - 2 )( x + y + 2 )

g) x² - 5x + 4

= x² - x - 4x + 4

= x( x - 1 ) - 4( x - 1 )

= ( x - 1 )( x - 4 )

h) x⁴ + 5x² + 4

= x⁴ + x² + 4x² + 4

= x²( x² + 1 ) + 4( x² + 1 )

= ( x² + 1 )( x² + 4 )

i) 2x² + 3x - 5

= 2x² - 5x + 2x - 5

= 2x( x + 1 ) - 5( x + 1 )

= ( x + 1 )( 2x - 5 )

k) x³ - 2x² + 6x - 5 ( không biết làm )
l) x² - 4x + 3

= ( x² - 4x + 4 ) - 1

= ( x - 2 )² - 1

= ( x - 3 )( x - 1 )

# Học tốt #

a: \(P=x^2+y^2-6x-2y+17\)

\(=x^2-6x+9+y^2-2y+1+7\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-1=0

=>x=3 và y=1

b: \(Q=x^2+xy+y^2-3x-3y+999\)

\(=x^2+x\left(y-3\right)+y^2-3y+999\)

\(=x^2+2\cdot x\cdot\left(\frac12y-\frac32\right)+\left(\frac12y-\frac32\right)^2+y^2-3y-\left(\frac12y-\frac32\right)^2+999\)

\(=\left(x+\frac12y-\frac32\right)^2+y^2-3y-\left(\frac14y^2-\frac32y+\frac94\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34y^2-\frac32y-\frac94+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y-3\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y+1-4\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y-1\right)^2+996\ge996\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x+\frac12y-\frac32=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=-\frac12y+\frac32=-\frac12+\frac32=\frac22=1\end{cases}\)

c: \(R=2x^2+2xy_{}+y^2-2x+2y+15\)

\(=x^2-4x+4+x^2+2xy+y^2+2x+2y+11\)

\(=\left(x-2\right)^2+x^2+2xy+y^2+2x+2y+1+10\)

\(=\left(x-2\right)^2+\left(x+y+1\right)^2+10\ge10\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y+1=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x-1=-2-1=-3\end{cases}\)

d: \(S=x^2+26y^2-10xy+14x-76y+59\)

\(=x^2-10xy+25y^2+14x-70y+y^2-6y+59\)

\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+y^2-6y+9+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-3=0\\ x-5y+7=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=5y-7=5\cdot3-7=15-7=8\end{cases}\)

e: \(T=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x-2y+5=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=2y-5=2\cdot1-5=2-5=-3\end{cases}\)

\(x^2-6x+y^2-2y+12\)

= \(\left(x^2-6x+9\right)+\left(y^2-2y+1\right)+2\)

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

mà \(\left(x-3\right)^2\) ≥ 0 ∀ x và \(\left(y-1\right)^2\)≥ 0 ∀ y

=> Min A = 2 <=> \(\left(x-3\right)^2\)=0 <=> x=3

và (y-1)\(^2\)=0 <=> y=1

vậy Min A =2 khi x=3 và y=1

Lần sau bạn chú ý viết đầy đủ yêu cầu của đề bài.