Giải các phương trình sau:

a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)

b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)

c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)

d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)

Bài 1 : Giải bất phương trình sau

1 , \(\left(2x+3\right)\left(5x-7\right)\ge0\)

2 , \(\left(3-2x\right)\left(4x+3\right)< 0\)

3 , \(\left(2x+5\right)\left(3-x\right)\left(5x-1\right)\le0\)

4 , \(x^2-3x+2< 0\)

5 , \(-x^2+12x+13>0\)

6 , \(x^2+6x+9\le0\)

7 , \(\frac{x+2}{3x+1}>\frac{x-2}{2x-1}\)

8 , \(\frac{1}{x+2}< \frac{3}{x-3}\)

9 , \(\frac{5x-6}{2x-5}\le6\)

10 , \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Giải phương trình:

1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)

4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)

5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)

6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)

8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)

9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)

10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)

2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)

4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

1) Giải các pt sau:

a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

b) \(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)

c) \(\frac{x+8}{6}-\frac{2x-5}{5}=\frac{x-1}{3}-x+7\)

d) \(\frac{7x}{8}-5\left(x-9\right)=\frac{2x+1,5}{6}\)

e) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

f) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)

Giải các phương trình,bất phương trình:

c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)

e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)

g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)

h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)

k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)

l,\(\left(x^2-2x+1\right)-4=0\)

m,\(4x^2+4x++1=x^2\)

Xin đáy ai giúp mình đi

Bài 1:tính giá trị biểu thức sau:

\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\) 

Bài 2 tìm x biết

\(a,1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)

\(b,\left|3x-\frac{2}{5}\right|=\frac{1}{35}+\left|\frac{-3}{7}\right|\)

\(c,-3\frac{1}{2}-\left(2x+\frac{1}{2}\right):\left(-\frac{3}{7}\right)=0\)

\(d,\left(x-4\right).\left(x+2\right)\le0\)với x thuộc z

giúp mk với

Tìm điều kiện xác định rồi giải các phương trình sau:

a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!