a)

b)

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

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

\(-2x>-3\)

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

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

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

\(2x^2-2x+1\leq2x^2+6x+5\)

\(-8x\leq4\)

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

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

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

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

\(-11x\leq-2\)

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

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

a) \(\frac{3x+5}{2}-x\geq1+\frac{x+2}{3}\)

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

\(9x+15-6x\geq6+2x+4\)

\(9x-6x-2x\geq6+4-15\)

\(x\geq-5\)

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

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

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

\(2x-4-6x-12\leq3x-51\)

\(-4x-16\leq3x-51\)

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

\(-7x\leq-35\)

\(x\geq5\)

Vậy nghiệm của bất phương trình đã cho là: \(x\geq5\)

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

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

\(8x+4-3x+12\leq6x+2-x+4\)

\(5x+16\leq5x+6\)

\(5x-5x\leq6-16\)

\(0x\leq-10\)

Vậy bất phương trình đã cho vô nghiệm.

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

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

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

\(0x>81\)

Vậy bất phương trình vô nghiệm.

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

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

\(12x-3+6x-19\leq18x-22\)

\(0x\leq0\)

Bất phương trình có nghiệm bất kì.

a) Điều kiện \(1-x\neq0\); \(1-2x\neq0\) và \(1+x\neq0\) hay \(x\neq1\); \(x\neq\frac{1}{2}\) và \(x\neq-1\)

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

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

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

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

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

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

\(2x-1<0\) vì \(- 2 < 0\)

\(x<\frac{1}{2}\) (nhận)

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