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

\(=\left(x^2-y^2\right)\cdot1+x^2\cdot1+3y^2=x^2-y^2+x^2+3y^2=2x^2+2y^2=2\left(x^2+y^2\right)=2\cdot1=2\)

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

\(-3x^4-3y^4=2\cdot1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)

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

1/

d/ \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

<=> \(24x^2+7x-6-\left(4x^2+23x+28\right)-\left(10x^2+3x-1\right)=-33\)

<=> \(24x^2+7x-6-4x^2-23x-28-10x^2-3x+1=-33\)

<=> \(10x^2-19x-33=-33\)

<=> \(10x^2-19x=0\)

<=> \(x\left(10x-19\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\10x-19=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{10}\end{cases}}\)

\(A=4x^2-12x+11\)

\(A=\left(2x\right)^2-2.2x.3+3^2+2\)

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

Ta có: \(\left(2x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-3\right)^2+2\ge2\forall x\)

Dấu = xảy ra \(\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Vậy A_min=2\(\Leftrightarrow x=\frac{3}{2}\)

\(B=x^2-2x+y^2+4y+6\)

\(B=\left(x^2-2x+1\right)+\left(y^2+2.2y+2^2\right)+1\)

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

Ta có:  \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\forall x;y}\)

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

Vậy B_min=1\(\Leftrightarrow x=1;y=-2\)

\(A=-x^2-6x+1\)

\(\Rightarrow-A=x^2+6x-1\)

\(-A=\left(x^2+2.3x+3^2\right)-10\)

\(-A=\left(x+3\right)^2-10\)

\(\Rightarrow A=-\left(x+3\right)^2+10\)

Ta có: \(\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\forall x\Rightarrow-\left(x+3\right)^2+10\le10\forall x\)

Dấu = xảy ra \(\Leftrightarrow-\left(x+3\right)^2=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy A_max=10\(\Leftrightarrow\)x= -3

Sửa đề:

\(B=-2x^2-8x-6\)

\(B=-2.\left(x^2+2.2x+2^2\right)+2\)

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

Ta có: \(2.\left(x+2\right)^2\ge0\forall x\Rightarrow-2.\left(x+2\right)^2\le0\forall x\Rightarrow-2.\left(x+2\right)^2+2\le2\forall x\)

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

Vậy B_max=2\(\Leftrightarrow x=-2\)

Đề phải là tìm min mới đúng

a, A=4x²-12x+11

=(4x²-12x+9)+2

=(2x-3)²+2

Vì (2x-3)² \(\ge\) 0 => A=(2x-3)²+2 \(\ge\) 2

Dấu "=" xảy ra khi 2x-3=0 <=> x=3/2

Vậy Amin = 2 khi x=3/2

b, B=x²-2x+y²+4y+6

=(x²-2x+1)+(y²+4y+4)+1

=(x-1)²+(y+2)²+1

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

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

\(\Rightarrow B=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu "=" xảy ra khi x=1,y=-2

Vậy Bmin = 1 khi x=1,y=-2

2   \(x^7+x^5+1=x^7+x^6+x^5-x^6+1=x^5\left(x^2+x+1\right)-\left(x^6-1\right)=x^5\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)

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

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

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

\(=\left(x^2-6x+9\right)\left(x+1\right)=\left(x-3\right)^2\left(x+1\right)\)

(​x+2)[(x+3)-(x+5)]=0

suy ra (x+2)(x-x+3-5)=0

-2(x+2)=0

suy ra x+2=0 suy ra x=2

a ) ( x + 2 )[ ( x + 3 ) - ( x + 5 ) ] = 0 

      ( x + 2 )( x - 2 ) = 0 

      x^2 - 4 = 0 

      x^2 = 4

      => x = 2