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

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

2x 3 -2bx-24 x+4x+3 2 2x - 2x + 8x 3 2 +6x -8x 2 -8b -24 -8 - -8x-32x -24 2 -8b + 32x

Bài dưới lỗi kĩ thuật => sửa :'(

x + 4x + 3 3 2 - 2x + 8x + 6x 2x - 2bx - 24 3 2 -8x - 8b - 24 - -8x - 32x - 24 -8b + 32x 2x - 8 2 2

Vậy ( 2x³ - 2bx - 24):( x² + 4x + 3) = 2x - 8 ( dư -8b + 32x )

Đây là bất đẳng thức Cauchy Schwarz và CM bằng cách biến đổi tương đương như sau:

Ta có: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+2xy+y^2-4xy\ge0\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: x = y

\(=x^2-4x+4-x^2-6x-9-4x-4\)

\(=-14x-9\)