a) ( 2x + 3 )² - 2( 2x + 3 )( 2x + 5 ) + ( 2x + 5 )²

= [ ( 2x + 3 ) - ( 2x + 5 ) ]²

= ( 2x + 3 - 2x - 5 )²

= (-2)² = 4

b) ( x² + x + 1 )( x² - x + 1 )( x² - 1 )

= ( x⁴ - x³ + x² + x³ - x² + x + x² - x + 1 )( x² - 1 )

= ( x⁴ + x² + 1 )( x² - 1 )

= x⁶ - x⁴ + x⁴ - x² + x² - 1

= x⁶ - 1 

c) ( x + y )² + ( x - y )²

= x² + 2xy + y² + x² - 2xy + y²

= 2x² + 2y² = 2( x² + y² )

d) 2( x - y )( x + y ) + ( x + y )² + ( x - y )²

= [ ( x + y ) + ( x - y ) ]²

= ( x + y + x - y )²

= ( 2x )² = 4x²

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

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

= [ ( x - y + z ) - ( z - y ) ]²

= ( x - y + z - z + y )²

= x²

f) ( a + b - c )² + ( a - b + c )² - 2( b - c )²

= [ ( a + b ) - c ]² + [ ( a - b ) + c ]² - 2( b² - 2bc + c² )

= [ ( a + b )² - 2( a + b )c + c² ] + [ ( a - b )² + 2( a - b )c + c² ] - 2b² + 4bc - 2c²

= a² + b² + c² + 2ab - 2bc - 2ca + c² + a² + b² + c² - 2ab + 2bc + 2ac - 2b² + 4bc - 2c²

= 2a² 

g) ( a + b + c )² + ( a - b - c )² + ( b - c - a )² + ( c - a - b )²

= [ ( a + b ) + c ]² + [ ( a - b ) - c ]² + [ ( b - c ) - a ]² + [ ( c - a ) - b ]²

= [ ( a + b )² + 2( a + b )c + c² ] + [ ( a - b )² - 2( a - b )c + c² ] + [ ( b - c )² - 2( b - c )a + a² ] + [ ( c - a )² - 2( c - a )b + b² ]

= [ a² + b² + c² + 2ab + 2bc + 2ca ] + [ a² + b² + c² - 2ab + 2bc - 2ca ] + [ a² + b² + c² - 2ab - 2bc + 2ca ] + [ a² + b² + c² + 2ab - 2bc - 2ca ]

= 4a² + 4b² + 4c²

Có vẻ hơi dài dòng nhỉ :( Nhưng như này là kĩ nhất đấy :)

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

dùng hđt \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

có nhân tử chung

a. \(A=100-2x-x^2=-\left(x+1\right)^2+101\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+101\le101\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy maxA = 101 <=> x = - 1

b. \(B=-3x^2+x=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)

Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\) \(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{1}{6}\right)^2=0\Leftrightarrow x=\frac{1}{6}\)

Vậy maxB = 1/12 <=> x = 1/6 

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\le\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

Vậy maxC = 3/4 <=> x = 1/2

A = 100 - 2x - x²

= -( x² + 2x + 1 ) + 101

= -( x + 1 )² + 101 ≤ 101 ∀ x

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MaxA = 101 <=> x = -1

B = -3x² + x

= -3( x² - 1/3x + 1/36 ) + 1/12

= -3( x - 1/6 ) + 1/12 ≤ 1/12 ∀ x

Đẳng thức xảy ra <=> x - 1/6 = 0 => x = 1/6

=> MaxB = 1/12 <=> x = 1/6

C = 3x( 1 - x )

= -3x² + 3x

= -3( x² - x + 1/4 ) + 3/4

= -3( x - 1/2 )² + 3/4  ≤ 3/4 ∀ x

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

=> MaxC = 3/4 <=> x = 1/2

\(D=12x-5-x^2=-\left(x-6\right)^2+31\)

Vì \(\left(x-6\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-6\right)^2+31\le31\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x-6\right)^2=0\Leftrightarrow x=6\)

Vậy maxD = 31 <=> x = 6

D = 12x - 5 - x²

= -( x² - 12x + 36 ) + 31

= -( x - 6 )² + 31 ≤ 31 ∀ x

Đẳng thức xảy ra <=> x - 6 = 0 => x = 6

=> MaxD = 31 <=> x = 6

A = 2x² + y² - 2xy - 2y + 2000 = (x² - 2xy + y²) + 2(x - y) + 1 + (x² + 2x + 1) + 1998

= (x - y)² + 2(x - y) + 1 + (x + 1)² + 1998 = (x - y + 1)² + (x + 1)² 1998 \(\ge\)1998 với mọi x,y

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

Vậy MinA = 1998 khi x = -1 và y = .0

b) B = x² + 5y² - 2xy + 6x - 18y + 50 = (x² - 2xy + y²) + 6(x - y) + 9 + (4y² - 12y + 9) + 32

= (x - y)² + 6(x - y) + 9 + (2y - 3)² + 32 = (x - y + 3)² + (2y - 3)² + 32 \(\ge\)32 với mọi x,y

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

Vậy MinB = 32 khi x = -3/2 và y = 3/2

c) C = 3x² +  x + 4 = 3(x² + 1/3x + 1/36) + 47/12 = 3(x + 1/6)² + 47/12 > = 47/12 với mọi x

Dấu "=" xảy ra <=> x + 1/6 = 0 <=> x = -1/6

Vậy MinC = 47/12 khi x = -1/6

A = 2y² + x² - 2xy - 2y + 2000 ( vầy mới tính được bạn nhé ;-; )

= ( x² - 2xy + y² ) + ( y² - 2y + 1 ) + 1999

= ( x - y )² + ( y - 1 )² + 1999

\(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y\right)^2+\left(y-1\right)^2+1999\ge1999\forall x,y\)

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

=> MinA = 1999 <=> x = y = 1

B = x² + 5y² - 2xy + 6x - 18y + 50

= ( x² - 2xy + y² + 2x - 6y + 9 ) + ( 4y² - 12y + 9 ) + 32

= [ ( x² - 2xy + y² ) + 2( x - y ).3 + 3² ] + ( 2y - 3 )² + 32

= [ ( x - y )² + 2( x - y ).3 + 3² ] + ( 2y - 3 )² + 32

= ( x - y + 3 ) + ( 2y - 3 )² + 32

\(\hept{\begin{cases}\left(x-y+3\right)^2\ge0\forall x,y\\\left(2y-3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y+3\right)^2+\left(2y-3\right)^2+32\ge32\forall x,y\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y+3=0\\2y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

=> MinB = 32 <=> x = -3/2 ; y = 3/2

C = 3x² + x + 4

= 3( x² + 1/3x + 1/36 ) + 47/12

= 3( x + 1/6 )² + 47/12 ≥ 47/12 ∀ x

Đẳng thức xảy ra <=> x + 1/6 = 0 => x = -1/6

=> MinC = 47/12 <=> x = -1/6