Tử \(x^4+2x^3+8x+16\)

\(=x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16\)

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

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

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

Mẫu \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

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

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

Thay tử và mẫu vào ta có:\(\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\)

Dấu "=" khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy Min=0 khi x=-2

a) ĐKXĐ: \(x\notin\left\{\frac{1}{2};\frac{-1}{2}\right\}\)

Ta có: \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)

\(\Leftrightarrow\frac{8x+1}{4\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}+\frac{8x^2}{3\left(4x^2-1\right)}\)

\(\Leftrightarrow\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}=\frac{2x\cdot4\cdot\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}+\frac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(3\left(8x+1\right)\left(2x-1\right)=8x\left(2x+1\right)+32x^2\)

\(\Leftrightarrow3\left(16x^2-8x+2x-1\right)=16x^2+8x+32x^2\)

\(\Leftrightarrow3\left(16x^2-6x-1\right)=48x^2+8x\)

\(\Leftrightarrow48x^2-18x-3-48x^2-8x=0\)

\(\Leftrightarrow-26x-3=0\)

\(\Leftrightarrow-26x=3\)

hay \(x=-\frac{3}{26}\)

Vậy: \(S=\left\{-\frac{3}{26}\right\}\)

b) Ta có: \(\left(x-2\right)\left(x-3\right)< \left(x-4\right)^2-2\left(x+3\right)\)

\(\Leftrightarrow x^2-5x+6< x^2-8x+16-2x-6\)

\(\Leftrightarrow x^2-5x+6< x^2-10x+10\)

\(\Leftrightarrow x^2-5x+6-x^2+10x-10< 0\)

\(\Leftrightarrow5x-4< 0\)

\(\Leftrightarrow5x< 4\)

hay \(x< \frac{4}{5}\)

Vậy: S={x|\(x< \frac{4}{5}\)}

\(\frac{2x+1}{x^2-5x+4}+\frac{5}{x-1}=\frac{2}{x-4}\)ĐKXĐ : \(x\ne1;4\)

\(\Leftrightarrow\frac{2x+1}{\left(x-1\right)\left(x-4\right)}+\frac{5\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)

\(\Leftrightarrow2x+1+5x-20=2x-2\)

\(\Leftrightarrow2x+5x-2x=-1+20-2\)

\(\Leftrightarrow5x=17\)

\(\Leftrightarrow x=\frac{17}{5}\)

KL : Nghiệm của PT là S={ 17/5 }

\(\frac{7}{8x}-\frac{x-5}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{7}{8x}-\frac{x-5}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{2\left(x-5\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

\(\Leftrightarrow7x-14-2x+10=4x-4+x\)

\(\Leftrightarrow7x-2x-4x-x=14-10-4\)

\(\Leftrightarrow0x=0\)

=> PT vô số nghiệm