Ta có : 

\(\frac{3x-2}{5}\ge\frac{x}{2}+0,8\)

\(\Leftrightarrow x\ge12\)

và \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)

\(\Leftrightarrow x< 13\)   \(x\in Z\)

\(\Rightarrow x=12\)

b, \(\frac{3x-2}{5}\ge\frac{x+1,6}{2}\)

=> \(6x-4\ge5x+8\)

=> \(x-12\ge0\)

=> \(x\ge12\)

bpt 2: \(\frac{6-2x+5}{6}>\frac{3-x}{4}\)

=> \(\frac{11-2x}{6}>\frac{3-x}{4}\)

=> \(44-8x>18-6x\)

=> \(x< 13\)

Vậy để t/m cả 2 bpt thì : \(12\le x< 13\)

a, \(\frac{x^2+x^2-4}{x\left(x-2\right)}>2\) (Đk : \(x\ne\left(0;2\right)\))

=> \(2x^2-4>2x^2-4x\)

=> \(4x-4=4\left(x-1\right)>0\)

=> \(x>1\)(t/m) 

\(\frac{3x-2}{5}\ge\frac{x}{2}+\frac{4}{5}\Leftrightarrow2\left(3x-2\right)\ge5x+8\)

\(\Leftrightarrow x\ge12\) (1)

\(1-\frac{2x-5}{6}>\frac{3-x}{4}\Leftrightarrow12-2\left(2x-5\right)>3\left(3-x\right)\)

\(\Leftrightarrow22-4x>9-3x\Leftrightarrow x< 13\) (2)

Từ (1) và (2) \(\Rightarrow12\le x< 13\)

Mà \(x\in Z\Rightarrow x=12\)

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

\(\Rightarrow\frac{12x}{30}+\frac{10\left(3-2x\right)}{30}-\frac{15\left(3x+2\right)}{30}\ge0\)

\(\Rightarrow12x+30-20x-45x-30\ge0\)

\(\Rightarrow-53x\ge0\)\(\Leftrightarrow x\le0\)\(\left(1\right)\)

\(\frac{x}{2}+\frac{3-2x}{5}\ge\frac{3x-5}{6}\)

\(\Rightarrow\frac{15x}{30}+\frac{6\left(3-2x\right)}{30}-\frac{5\left(3x-5\right)}{30}\ge0\)

\(\Rightarrow15x+18-12x-15x+25\ge0\)

\(\Rightarrow-12x\ge-43\)\(\Rightarrow12x\le43\Leftrightarrow x\le\frac{43}{12}\)\(\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có tập nghiệm chung của cả hai phương trình là \(x\le0\)

Bài 1:

d)ĐKXĐ: \(x\ne8\)

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

MTC=24(x-8)

\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)

\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)

\(\Leftrightarrow348-29x=0\)

\(\Leftrightarrow-29x+348=0\)

\(\Leftrightarrow x=\frac{-348}{-29}=12\)

Vậy: x=12

e) ĐKXĐ: \(x\ne\pm1\)

Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)

MTC=4(x+1)(x-1)

\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)

\(\Leftrightarrow20x+4=0\)

\(\Leftrightarrow20x=-4\)

\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)

Vậy: x không có giá trị

g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)

\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)

MTC=2(x+1)

\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)

\(\Leftrightarrow2-x+1=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)

Vậy: x không có giá trị

Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà