a) \(\frac{3 \left(\right. 2 x + 1 \left.\right)}{20} + 1 > \frac{3 x + 52}{10}\)

\(\frac{3 \left(\right. 2 x + 1 \left.\right)}{20} + \frac{20}{20} > \frac{2 \left(\right. 3 x + 52 \left.\right)}{20}\)

\(6 x + 3 + 20 > 6 x + 104\)

\(0 x > 81\)

Vậy bất phương trình vô nghiệm.

b) \(\frac{4 x - 1}{2} + \frac{6 x - 19}{6} \leq \frac{9 x - 11}{3}\)

\(\frac{3 \left(\right. 4 x - 1 \left.\right)}{6} + \frac{6 x - 19}{6} \leq \frac{2 \left(\right. 9 x - 11 \left.\right)}{6}\)

\(12 x - 3 + 6 x - 19 \leq 18 x - 22\)

\(0 x \leq 0\)

Bất phương trình có nghiệm bất kì.

a) \(\frac{3 x + 5}{2} - x \geq 1 + \frac{x + 2}{3}\)

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

\(9 x + 15 - 6 x \geq 6 + 2 x + 4\)

\(9 x - 6 x - 2 x \geq 6 + 4 - 15\)

\(x \geq - 5\)

Vậy nghiệm của bất phương trình đã cho là: \(x \geq - 5\)

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

\(\frac{2 \left(\right. x - 2 \left.\right) - 6 x - 6.2}{6} \leq \frac{3 \left(\right. x - 17 \left.\right)}{6}\)

\(2 x - 4 - 6 x - 12 \leq 3 x - 51\)

\(- 4 x - 16 \leq 3 x - 51\)

\(- 4 x - 3 x \leq - 51 + 16\)

\(- 7 x \leq - 35\)

\(x \geq 5\)

Vậy nghiệm của bất phương trình đã cho là: \(x \geq 5\)

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

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

\(8 x + 4 - 3 x + 12 \leq 6 x + 2 - x + 4\)

\(5 x + 16 \leq 5 x + 6\)

\(5 x - 5 x \leq 6 - 16\)

\(0 x \leq - 10\)

Vậy bất phương trình đã cho vô nghiệm.

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

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

\(- 2 x > - 3\)

\(x < \frac{3}{2}\)

Vậy nghiệm của bất phương trình đã cho là: \(x < \frac{3}{2}\)

b) \(\left(\left(\right. x - 1 \left.\right)\right)^{2} + x^{2} \leq \left(\left(\right. x + 1 \left.\right)\right)^{2} + \left(\left(\right. x + 2 \left.\right)\right)^{2}\)

\(2 x^{2} - 2 x + 1 \leq 2 x^{2} + 6 x + 5\)

\(- 8 x \leq 4\)

\(x \geq - \frac{1}{2}\)

Vậy nghiệm của bất phương trình đã cho là: \(x \geq - \frac{1}{2}\)

c) \(\left(\right. x^{2} + 1 \left.\right) \left(\right. x - 6 \left.\right) \leq \left(\left(\right. x - 2 \left.\right)\right)^{3}\)

\(x^{3} - 6 x^{2} + x - 6 \leq x^{3} - 6 x^{2} + 12 x - 8\)

\(- 11 x \leq - 2\)

\(x \geq \frac{2}{11}\)

Vậy nghiệm của bất phương trình đã cho là: \(x \geq \frac{2}{11}\).

) Điều kiện \(1 - x \neq 0\); \(1 - 2 x \neq 0\) và \(1 + x \neq 0\) hay \(x \neq 1\); \(x \neq \frac{1}{2}\) và \(x \neq - 1\)

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

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

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

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

\(A = \left[\right. \frac{- 2}{\left(\right. 1 - x \left.\right) \left(\right. 1 + x \left.\right)} \left]\right. . \frac{\left(\right. 1 - x \left.\right) \left(\right. 1 + x \left.\right)}{2 x - 1} = \frac{- 2}{2 x - 1}\)

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

\(2 x - 1 < 0\) vì \(- 2 < 0\)

\(x < \frac{1}{2}\) (nhận)

Vậy \(x < \frac{1}{2}\) và \(x\neq-1\) thì \(A > 0\).

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