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) H=x² - 4x +16

<=> H=x² -4x + 4 + 12

<=> H=(x-2)² +12 \(\ge12\)

Vậy Min H = 12

Dấu "=" xảy ra khi x=2

\(K=x^2-6xy+9y^2+4\left(x-3y\right)+4+x^2-12x+36+1978\)

\(K=\left(x-3y\right)^2+4\left(x-3y\right)+2^2+\left(x-6\right)^2+1978\)

\(K=\left(x-3y+2\right)^2+\left(x-6\right)^2+1978\ge1978\)

Vậy Min K =1978

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-3y+2=0\\x-6=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=\dfrac{8}{3}\\x=6\end{matrix}\right.\)

Cái phần cuối câu K nó bị gì đó , nếu không thấy thì bạn tự giải tiếp nhen

A=2x^2+9y^2-6xy-6x-12y+2024 
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2024
A = (x -3y)^2 +4(x -3y) + 4 + x^2 -10x +25 + 1995
A = (x -3y +2)^2 + (x -5)^2 + 1995 \geq 1995
Min A = 1995 
 x - 5 = 0 => x = 5
Và x - 3y + 2 = 0 hay 5 -3y +2 = 0 => -3y = -7 => y = 7/3 

\(K\)\(nha!~!\)

C = x² - 4x + 16
   = (x² - 4x + 4) + 12
   = (x - 2)² + 12

Vậy C_min = 12 (vì \(\left(x-2\right)^2\ge0\Leftrightarrow\left(x-2\right)^2+12\ge12\))

Còn D mình không biết cách làm

Thôi em làm luôn nha:)

\(D=\left(x^2-2.x.3y+9y^2\right)+4\left(x-3y\right)+4+x^2-2.x.6+36+1978\)

\(=\left(x-3y\right)^2+2\left(x-3y\right).2+2^2+\left(x-6\right)^2+1978\)

\(=\left(x-3y+2\right)^2+\left(x-6\right)^2+1978\ge1978\)

Đẳng thức xảy ra x =6, y = 8/3

\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)

Nỗi hứng lm cho vui!

Bài 1:

a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)

Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12

=> Dấu = xảy ra <=> \(x=2\)

b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)

= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)

= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)

= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)

Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)

Bài 2:

a) P = \(-x^2-4x+16=-\left(x^2+4x+4\right)+20\)

= \(-\left(x+2\right)^2+20\le20\)

=> Dấu = xảy ra <=> \(x=-2\)

b) \(Q=-x^2+2xy-4y^2+2x+10y-2017\)

= \(-\left[\left(x^2-2xy+y^2\right)+3\left(y^2-4y+4\right)-2\left(x-y\right)+2005\right]\)

= \(-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2004\right]\)

= \(-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]-2004\)

Vì \(\left\{{}\begin{matrix}-\left(x-y-1\right)^2\le0\\3\left(y-2\right)^2\le0\end{matrix}\right.\) => Q \(\le-2004\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)

đăng từng câu 1 thôi

a)\(x^4+3x^3+x^2+3x=x\left(x^3+3x^2+x+3\right)\)

\(=x\left[x^2\left(x+3\right)+\left(x+3\right)\right]=x\left(x+3\right)\left(x^2+1\right)\)

b) \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-4z^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)

c) \(=2x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(2x-7\right)\)

\(a,=x^3\left(x+3\right)+x\left(x+3\right)=x\left(x^2+1\right)\left(x+3\right)\\ b,=\left(x+3y\right)^2-4z^2=\left(x+3y+2z\right)\left(x+3y-2z\right)\\ c,=2x^2-2x-7x+7=\left(x-1\right)\left(2x-7\right)\)

\(a)=x^3(x+3)+x(x+3)=(x^2+x)(x+3)=x(x+1)(x+3)\\b)=(x+3y)^2-4z^2=(x+3y-2z)(x+3y+2z)\\c)=2x^2-2x-7x+7=2x(x-1)-7(x-1)=(2x-7)(x-1)\)

\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)

\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)