\(A,B,C,D\inℤ\)

\(f\left(x\right)-g\left(x\right)\)

\(=x^{2n}-x^{2n-1}+...+x^2-x+1+x^{2n+1}-x^{2n}+x^{2n-1}-...-x^2+x-1\)

\(=x^{2n+1}\)

\(=\left(\dfrac{1}{10}\right)^{2n+1}=\dfrac{1}{10^{2n+1}}\)

a: \(x^2+4x-1=0\)

=>\(x^2+4x+4-5=0\)

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

=>\(\left[\begin{array}{l}x+2=\sqrt5\\ x+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt5-2\\ x=-\sqrt5-2\end{array}\right.\)

b: \(2x^2-4x+1=0\)

=>\(2\left(x^2-2x+\frac12\right)=0\)

=>\(x^2-2x+\frac12=0\)

=>\(x^2-2x+1-\frac12=0\)

=>\(\left(x-1\right)^2=\frac12\)

=>\(\left[\begin{array}{l}x-1=\frac{\sqrt2}{2}\\ x-1=-\frac{\sqrt2}{2}\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=\frac{\sqrt2+2}{2}\\ x=\frac{-\sqrt2+2}{2}\end{array}\right.\)

c: \(\left(2x-1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2x-9\)

=>\(2x^2+4x-x-2-\left(x^2-3x+2\right)-2x+9=0\)

=>\(2x^2+x+7-x^2+3x-2=0\)

=>\(x^2+4x+5=0\)

=>\(x^2+4x+4+1=0\)

=>\(\left(x+2\right)^2+1=0\) (vô lý)

=>Phương trình vô nghiệm

d: \(\left(x-1\right)\cdot x\cdot\left(x+1\right)\left(x+2\right)=8\)

=>\(x\left(x+1\right)\left(x+2\right)\left(x-1\right)=8\)

=>\(\left(x^2+x\right)\left(x^2+x-2\right)=8\)

=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-8=0\)

=>\(\left(x^2+x-4\right)\left(x^2+x+2\right)=0\)

mà \(x^2+x+2=x^2+x+\frac14+\frac74=\left(x+\frac12\right)^2+\frac74\ge\frac74>0\forall x\)

nên \(x^2+x-4=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-4\right)=1+16=17>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-1-\sqrt{17}}{2\cdot1}=\frac{-1-\sqrt{17}}{2}\\ x=\frac{-1+\sqrt{17}}{2\cdot1}=\frac{-1+\sqrt{17}}{2}\end{array}\right.\)

dddddddddddddddddd

1)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2+x^2-6xy+9y^2-6x-12y+2004\)

\(=x^2+\left(x-3y\right)^2-10x+4x-12y+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

Dấu "=" khi \(\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

Vậy Min=1975 khi \(\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

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

Đặt \(t=x^2+x\) ta có:

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

\(=\left(t-2\right)^2-4\ge-4\)

Dấu "=" khi \(t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

Vậy Min=-4 khi \(\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

3)\(\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]\)

\(=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]\)

Đặt \(t=x^2+5x+5\) ta có:

\(t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" khi \(t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}\)\(\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

Vậy Min=\(-\frac{1}{4}\) khi \(x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

4)\(\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)

\(=\left(x^2-4x+3\right)\left(x^2-4x+5\right)\)

Đặt \(t=x^2-4x+3\) ta có:

\(t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)

Dấu "=" khi \(t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2\)

Vậy Min=-1 khi x=2

Thank you !

Bài này khá ez thôi: 

a) bạn sửa lại đề rồi làm theo cách làm của b,c,d nhé

b) Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)

\(\Rightarrow5x\ge0\Rightarrow x\ge0\) khi đó:

\(PT\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow x=5\)

c,d tương tự nhé

c,\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}+\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\forall x\)

\(\Rightarrow50x\ge0\Rightarrow x\ge0\)Khi đó:

\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)

\(\Rightarrow49x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=50x\)

\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{49}{99}\)