ĐKXĐ: 
x∉{1;−1;12}x∈{1;−1;\(\frac12\) ​}\(\)

\(A = \left(\right. \frac{1}{1 - x} + \frac{2}{x + 1} - \frac{5 - x}{1 - x^{2}} \left.\right) : \frac{1 - 2 x}{x^{2} - 1}\)

\(= \left(\right. \frac{- 1}{x - 1} + \frac{2}{x + 1} - \frac{x - 5}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} \left.\right) \cdot \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{- 2 x + 1}\)

\(= \frac{- \left(\right. x + 1 \left.\right) + 2 \left(\right. x - 1 \left.\right) - x + 5}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} \cdot \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{- 2 x + 1}\)

\(= \frac{- x - 1 + 2 x - 2 - x + 5}{- 2 x + 1} = \frac{2}{- 2 x + 1}\)

b: Để A>0 thì \(\frac{2}{- 2 x + 1} > 0\)

mà 2>0

nên -2x+1>0

=>-2x>-1

=>\(x < \frac{1}{2}\)

Kết hợp ĐKXĐ, ta được:  \(\begin{cases}x<\frac12\\ xkhác-1\end{cases}\)\(\)

a, \(x^{2} - 3 x + 1 > 2 \left(\right. x - 1 \left.\right) - x \left(\right. 3 - x \left.\right)\)

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

=>-3x+1>-x-2

=>-2x>-3

=>

b: \(\left.\left(\right.x-1\right)^2+x^2<=\left(x+1\right)^2+\left(x+2\right)^2\)

=>\(x^{2} - 2 x + 1 + x^{2} < = x^{2} + 2 x + 1 + x^{2} + 4 x + 4\)

=>-2x+1<=6x+5

=>-7x<=4

=>\(x > = - \frac{4}{7}\)

c: 

\(\left(\right.x^2+1\left.\right)\left(\right.x-6\left.\right)<=\left(x-2\right)^3\)

=>\(x^{3} - 6 x^{2} + x - 6 < = x^{3} - 6 x^{2} + 12 x - 8\)

=>x-6<=12x-8

=>-11x<=-8+6=-2

=>\(x > = \frac{2}{11}\)\(\)

=>\(\frac{3 x + 5 - 2 x}{2} > = \frac{3 + x + 2}{3}\)

=>\(\frac{x + 5}{2} - \frac{x + 5}{3} > = 0\)

=>\(\frac{3 \left(\right. x + 5 \left.\right) - 2 \left(\right. x + 5 \left.\right)}{6} > = 0\)

=>\(\frac{x + 5}{6} > = 0\)

=>x+5>=0

=>x>=-5

b,=>\(\frac{2 \left(\right. x - 2 \left.\right)}{6} + \frac{6 \left(\right. - x - 2 \left.\right)}{6} < = \frac{3 \left(\right. x - 17 \left.\right)}{6}\)

=>\(2 \left(\right. x - 2 \left.\right) + 6 \left(\right. - x - 2 \left.\right) < = 3 \left(\right. x - 17 \left.\right)\)

=>\(2 x - 4 - 6 x - 12 < = 3 x - 51\)

=>-4x-16<=3x-51

=>-7x<=-35

=>x>=5

c,=>\(\frac{4 \left(\right. 2 x + 1 \left.\right) - 3 \left(\right. x - 4 \left.\right)}{12} < = \frac{2 \left(\right. 3 x + 1 \left.\right) - x + 4}{12}\)

=>4(2x+1)-3(x-4)<=2(3x+1)-x+4

=>8x+4-3x+12<=6x+2-x+4

=>5x+16<=5x+6

=>16<=6(sai)

Vậy: BPT vô nghiệm

=>\(\frac{6 x + 3}{20} + \frac{20}{20} > \frac{6 x + 104}{20}\)

=>6x+23>6x+104

=>23>104(sai)

vậy: \(x \in \emptyset\)

b,=>\(\frac{3 \left(\right. 4 x - 1 \left.\right) + 6 x - 19}{6} < = \frac{2 \left(\right. 9 x - 11 \left.\right)}{6}\)

=>12x-3+6x-19<=18x-22

=>-22<=-22(luôn đúng)

Vậy: \(x \in R\)\(\)