Điều kiện: \(x\ne1\)

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{3x^2\left(x-1\right)-4x\left(x-1\right)+\left(x-1\right)}{2x^2\left(x-1\right)+x\left(x-1\right)-3\left(x-1\right)}\)

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

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

\(=\frac{3x\left(x-1\right)-\left(x-1\right)}{2x\left(x-1\right)+3\left(x-1\right)}\)

\(=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(2x+3\right)}=\frac{3x-1}{2x+3}\)

\(x^2-4x+1=x^2-2\cdot x\cdot2+4-4+1=\left(x-2\right)^2-4+1\)

\(=\left(x-2\right)^2-3\)    \(\forall x\in Z\)

\(\Rightarrow A_{min}=-3khix=2\)

\(a,A=x^2-4x+1=x^2-2.2.x+2^2-3=\left(x-2\right)^2-3\ge-3\)

dấu = xảy ra khi x-2=0

=> x=2

Vậy MinA=-3 khi x=2

\(b,B=5-8x-x^2=-\left(x^2+8x+5\right)=-\left(x^2+2.4.x+4^2\right)+9=-\left(x+4\right)^2+9\le9\)

dấu = xảy ra khi x+4=0

=> x=-4

Vậy MaxB=9 khi x=-4

\(c,C=5x-x^2=-\left(x^2-5x\right)=-\left(x^2-\frac{2.x.5}{2}+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

dấu = xảy ra khi \(x-\frac{5}{2}=0\)

=> x=\(\frac{5}{2}\)

Vậy Max C=\(\frac{25}{4}\)khi x=\(\frac{5}{2}\)

\(E=\frac{1}{x^2+5x+14}=\frac{1}{x^2+\frac{2.x.5}{2}+\frac{25}{4}+\frac{31}{4}}=\frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{31}{4}}\)

\(\left(x+\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)

dấu = xảy ra khi \(x+\frac{5}{2}=0\)

=> x\(=-\frac{5}{2}\)

vì tử thức >0,mẫu thức nhỏ nhất và lớn hơn 0 => E lớnnhất khi mẫu thức nhỏ nhất 

Vậy \(MaxE=\frac{31}{4}\)khi x\(=-\frac{5}{2}\)

ta có

=\(\frac{1!}{1!2}+\frac{2!2}{2!3!}+\frac{3!3}{3!4!}+...+\frac{6!6}{n!\left(n+1\right)!}=\frac{5039}{5040}->n=6\)

-- là giề

a(b^3-c^3) +b(c^3-a^3)+c(a^3-b^3)

=> a(b-c)(b^2+bc+c^2)+bc^3-ba^3+ca^3-cb^3

=>a(b-c)(b^2+bc+c^2)-(cb^3-bc^3)-(ba^3-ca^3)

=>a(b-c)(b^2+bc+c^2)-bc(b-c)(b+c)-a^3(b-c)

=>(b-c)(ab^2+abc+ac^2-cb^2-bc^2-a^3)

=>(b-c)(