\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow2+\frac{x+4}{2000}+\frac{x+3}{2001}=2+\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2001}+1\right)\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

Suy ra x+2004=0

\(\Leftrightarrow x=-2004\)

123456

Vì P(x) có hệ số bậc cao nhất là 1

Nên P(x) có thể được viết dưới dạng: \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

Và \(P\left(-1\right)=\left(-1\right)^5-5\left(-1\right)^3+4\left(-1\right)+1=1\)

\(P\left(\frac{1}{2}\right)=\frac{77}{32}\)

Ta có: \(Q\left(x\right)=2x^2+x-1=2x^2+2x-x-1=2x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(2x-1\right)\)

=> \(Q\left(x_1\right).\text{​​}\text{​​}Q\left(x_2\right).\text{​​}\text{​​}Q\left(x_3\right).\text{​​}\text{​​}Q\left(x_4\right).\text{​​}\text{​​}Q\left(x_5\right)\text{​​}\text{​​}\)

\(=\left(x_1+1\right)\left(2x_1-1\right)\left(x_2+1\right)\left(2x_2-1\right)\left(x_3+1\right)\left(2x_3-1\right)\left(x_4+1\right)\left(2x_4-1\right)\left(x_5+1\right)\left(2x_5-1\right)\)

\(=32\left(-1-x_1\right)\left(\frac{1}{2}-x_1\right)\left(-1-x_2\right)\left(\frac{1}{2}-x_2\right)\left(-1-x_3\right)\left(\frac{1}{2}-x_3\right)\left(-1-x_4\right)\left(\frac{1}{2}-x_4\right)\left(-1-x_5\right)\left(\frac{1}{2}-x_5\right)\)\(=32.P\left(-1\right).P\left(\frac{1}{2}\right)=32.1.\frac{77}{32}=77\)

\(p\left(x\right)=x^5-5x^3+4x+1=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

\(Q\left(x\right)=2\left(\frac{1}{2}-x\right)\left(-1-x\right)\)

Do đó \(Q\left(x_1\right)\cdot Q\left(x_2\right)\cdot Q\left(x_3\right)\cdot Q\left(x_4\right)\cdot Q\left(x_5\right)\)

\(=2^5\left[\left(\frac{1}{2}-x_1\right)\left(\frac{1}{2}-x_2\right)\left(\frac{1}{2}-x_3\right)\left(\frac{1}{2}-x_4\right)\left(\frac{1}{2}-x_5\right)\right]\)

\(=\left(-1-x_1\right)\left(-1-x_2\right)\left(-1-x_3\right)\left(-1-x_4\right)\left(-1-x_5\right)\)

\(=32P\left(\frac{1}{2}\right)\cdot\left[P\left(-1\right)\right]\)

\(=32\cdot\left(\frac{1}{32}-\frac{5}{8}+\frac{4}{2}+1\right)\left(-1+5-4+1\right)\)

\(=4300\)

*Mình không chắc*

Ta có hằng đẳng thức: 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Ta thấy \(\left(x-1\right)+\left(x-2\right)+\left(3-2x\right)=0\)

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

suy ra \(\left(x-1\right)\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{3}{2}\end{cases}}\)

\(S=\frac{29}{4}\).