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 

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

\(=\left(2x+1\right)\left[\left(2x\right)^2-2x.1+1-4\right]\)

\(=8x^3+1-4\left(2x+1\right)\)

\(=8x^3+1-8x-4=8x^3-8x-3\)

Ta có : 

\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ba}{c^2}\)

\(\Leftrightarrow P=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)

\(\Leftrightarrow P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)( 1 )

Biến đổi \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)ta được :

\(\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=\frac{-1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c^3}=0\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{-1}{c}\right)=0\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)( 2 )

Thay ( 2 ) vào ( 1 ) ta được

\(P=abc.\frac{3}{abc}=3\)