Ta có: \(a^2-b^2-c^2+2bc=a^2-\left(b^2+c^2-2bc\right)\)

\(=a^2-\left(b-c\right)^2=\left[a-\left(b-c\right)\right].\left[a+\left(b-c\right)\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\)

Vì a, b , c là các cạnh của tam giác 

\(\Rightarrow a+c>b\)\(\Rightarrow a+c-b>0\)

\(a+b>c\)\(\Rightarrow a+b-c>0\)

\(\Rightarrow\left(a+c-b\right)\left(a+b-c\right)>0\)

hay \(a^2-b^2-c^2+2bc>0\)( đpcm )

\(a^2-b^2-c^2+2bc=\)\(a^2-\left(b^2-2bc+c^2\right)\)\(=a^2-\left(b-c\right)^2\)\(=\left(a-b+c\right)\left(a+b-c\right)\)

\(\left(a+c-b\right)\left(a+b-c\right)\left(1\right)\)

Ta có theo bất đẳng thức tam giác thì: độ dài 1 cạnh bao giờ cũng nhỏ hơn tổng 2 cạnh còn lại

\(a+c>b\Rightarrow a+c-b>0\left(2\right)\)

\(a+b>c\Rightarrow a+b-c>0\left(3\right)\)

Từ (1) (2) (3) => \(\left(a+c-b\right)\left(a+b-c\right)>0\)

                          \(\Rightarrow a^2-b^2-c^2+2bc>0\)

a, \(\left(x^3-8\right):\left(x+2\right)\)

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

b, \(\left(x^3-3x^2+3x-1\right):\left(x-1\right)\)

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

x khác -2;4

\(M=\frac{\left(x^5-2x^4\right)+\left(2x^3-4x^2\right)-\left(3x-6\right)}{x^2-4x+2x-8}=\frac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x\left(x-4\right)+2\left(x-4\right)}\)

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

=0 khi x=1;-1;2

ko rút gọn đc

Ta có \(2x^3+9x^2+15x+9=M\left(2x+3\right)< =>M=\frac{2x^3+9x^2+15x+9}{2x+3}\)

Xét đa thức \(f\left(x\right)=2x^3+9x^2+15x+9=\left(2x^3+3x^2\right)+\left(6x^2+9x\right)+6x+9\)

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

Suy ra \(M=\frac{2x^3+9x^2+15x+9}{2x+3}=\frac{\left(2x+3\right)\left(x^2+3x+3\right)}{2x+3}=x^2+3x+3\)

Vậy đa thức \(M=x^2+3x+3\)

x^4 - x^3 + 6x^2 - x + a x^2 - x + 5 x^2 + 1 x^4 - x^3 + 5x^2 x^2 - x + a x^2 - x + 5 a-5

Để A chia hết B <=> a - 5 =0 <=> a = 5 

x^3 + 5x^2 + 7x - 13 x^2 + 6x + 13 x - 1 x^3 + 6x^2 + 13x -x^2 - 6x - 13 -x^2 - 6x - 13 0

Vậy : \(\left(x^3+5x^2+7x-13\right):\left(x^2+6x+13\right)=x-1\)

Sửa đề a, \(A=x^2+10x+196=\left(x^2+10x+25\right)+171\)

\(=\left(x+5\right)^2+171\)

Mà \(\left(x+5\right)^2\ge0\forall x;\left(x+5\right)^2+171\ge171\)

Vậy GTNN A = 171 <=> x = -5 

b, \(B=\left(x+1\right)^2+\left(3x-4\right)^2=x^2+2x+1+9x^2-24x+16\)

\(=10x^2-22x+17=10\left(x^2-2.\frac{11}{10}x+\frac{121}{100}\right)+\frac{49}{10}\)

\(=10\left(x-\frac{11}{10}\right)^2+\frac{49}{10}\)

Mà \(\left(x-\frac{11}{10}\right)^2\ge0\forall x;10\left(x-\frac{11}{10}\right)^2+\frac{49}{10}\ge\frac{49}{10}\)

Vậy GTNN B = 49/10 <=> x = 11/10