Ta có : n^3 - n^2 + n - 1 = n^2(n - 1) + (n - 1) = (n^2 + 1)(n - 1).
Để n^3 - n^2 + n - 1 là số nguyên tố thì ta có 2 TH :
TH1 : n^2 + 1 = 1 ; n - 1 nguyên tố => không có n thỏa mãn.
TH2 : n^2 + 1 nguyên tố, n - 1 = 1 => n = 2 (chọn)
Vậy n = 2 để n^3 - n^2 + n - 1 nguyên tố

Đặt A=x^4-x^3+3x^2-2x+2

=(x^4+3x^2+2)-(x^3+2x)

=(x^4+x^2+2x^2+2)-x(x^2+2)

=(x^2+1)(x^2+2)-x(x^2+2)

=(x^2+2)(x^2-x+1)

Ta có x^2+2>=2>0;

x^2-x+1=(x^2-x+1/4)+3/4 =(x-1/2)^2+3/4>=3/4>0 

=> A>0  

1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0

Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)

b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)

B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B =  \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{x+3y}{x\left(x-3y\right)}\)

\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)

\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)

Vậy (2x⁴-3x³+5x²-4x+3):(x²-x+1) = 2x² -x + 2 dư -x+1

Giả sử:

\(a>b>c\Rightarrow a-b>0,b-c>0,a-c>0\)

Ta có:

\(\hept{\begin{cases}a^2+b^2+c^2\ge a^2+c^2\\\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}\ge\frac{\left(\frac{1}{a-b}+\frac{1}{b-c}\right)^2}{2}\ge\frac{8}{\left(a-c\right)^2}\end{cases}}\)

Từ đây ta có:

\(VT\ge\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\)

Ta chứng minh

\(\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\ge\frac{9}{2}\)

\(\Leftrightarrow\left(a+c\right)^2\ge0\)(Đúng)

Vậy ta có điều phải chứng minh là đúng. Dấu = xảy ra khi a = - c; b = 0 và các hoán vị của nó

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

\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{b^2}{c+a}+\frac{bc}{a+b}\)

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

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}\)

\(+\frac{b\left(c+a\right)}{c+a}=a+b+c\)

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\left(a+b+c\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\left(đpcm\right)\)

hay

mày làm đi