a)ĐKXĐ: \(x\ne{\frac{1}{2}\left.\right.};x\ne1;x\ne-1\)

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

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

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

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

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: \(\begin{cases}x<\frac12\\ x\ne-1\end{cases}\) \(\)

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

=>\(x < \frac{3}{2}\)
Vậy \(x<\frac32\)
b)\(\left.\left(\right.x-1\right)^2+x^2\le\left.\left(\right.x+1\right)^2+\left.\left(\right.x+2\right)^2\)

=>\(x^2-2x+1+x^2\le x^2+2x+1+x^2+4x+4\)

=>-2x+1\(\le\) 6x+5

=>-7x\(\le\) 4

=>\(x\ge-\frac{4}{7}\)
Vậy \(x\ge-\frac47\)
c) \(\left(\right.x^2+1\left.\right)\left(\right.x-6\left.\right)\le\left(x-2\right)^3\)

=>\(x^3-6x^2+x-6\le x^3-6x^2+12x-8\)

=>x-6\(\le\) 12x-8

=>-11x\(\le\) -8+6

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

a: \(\frac{3 x + 5}{2}-x\ge1+\frac{x + 2}{3}\)

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

=>\(\frac{x + 5}{2}-\frac{x + 5}{3}\ge0\)

=>\(\frac{3 \left(\right. x + 5 \left.\right) - 2 \left(\right. x + 5 \left.\right)}{6}\ge0\)

=>\(\frac{x + 5}{6}\ge0\)

\(=>x+5\ge0\)

=>\(x\ge-5\)
Vậy bất phương trình có nghiệm \(x\ge-5\)

b: \(\frac{x - 2}{3}-x-2\le\frac{x - 17}{2}\)

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

=>\(2\left(\right.x-2\left.\right)+6\left(\right.-x-2\left.\right)\le3\left(\right.x-17\left.\right)\)

=>\(2x-4-6x-12\le3x-51\)

=>-4x-16\(\le\) 3x-51

=>-7x\(\le\) -35

=>x\(\ge\) 5
Vậy bất phương trình có nghiệm là x \(\ge\) 5

c: \(\frac{2 x + 1}{3}-\frac{x - 4}{4}\le\frac{3 x + 1}{6}-\frac{x - 4}{12}\)

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

=>4(2x+1)-3(x-4)\(\le\) 2(3x+1)-x+4

=>8x+4-3x+12\(\le\) 6x+2-x+4

=>5x+16\(\le\) 5x+6

=>16\(\le\) 6(sai)

Vậy bất phương trình vô nghiệm

a: \(\frac{3 x + 5}{2}-x\ge1+\frac{x + 2}{3}\)

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

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

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

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

=>x+5>=0

=>x>=-5

b: \(\frac{x - 2}{3} - x - 2 < = \frac{x - 17}{2}\)

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

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

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

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

=>-7x<=-35

=>x>=5

c: \(\frac{2 x + 1}{3} - \frac{x - 4}{4} < = \frac{3 x + 1}{6} - \frac{x - 4}{12}\)

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

=>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

My favorite community helper is the garbage collector. He has a strong body because his work is very hard. He usually wears an orange uniform and gloves to protect himself. He is friendly, hardworking and never complains about his job. Every day, he collects trash from the streets and houses to keep the environment clean and safe. Thanks to him, people can live in a healthier and more beautiful neighborhood. I respect him very much because he works early in the morning and even in the hot sun or the rain. He is a true hero in our community