Giải phương trình:

1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)

2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)

4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)

5.\(6x^4-5x^3-38x^2-5x+6=0\)

1) \(\frac{2}{5}x\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}x\frac{1}{3}\)

2)\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)

3)\(\left[6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right]:\frac{3}{12}\)

4)\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)

5)\(\left(-2\right)^3x\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)

6)\(\left(\frac{2}{5}\right)^2+5\frac{1}{2}x\left(45-2\right)+\frac{23}{4}\)

7)\(\frac{4}{9}-19\frac{1}{3}-\frac{4}{9}x39\frac{1}{3}\)

8)\(\left(-\frac{1}{2}\right)^2:\frac{1}{4}-2x\left(-\frac{1}{2}\right)^2\)

9)\(125\%x\left(-\frac{1}{2}\right)^3:\left(1\frac{5}{16}-1,5\right)+2008^0\)

giúp mình nha mai mình học rùi

1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )

\(a)\frac{-6}{11}.\frac{5}{13}+\frac{-6}{11}.\frac{8}{13}-\left(\frac{-2}{5}\right)^0\) \(b)\left(2\frac{2}{3}+3\frac{1}{2}\right);\left(4\frac{3}{4}-2\frac{1}{6}\right)+\frac{19}{31}\) \(c)2,4:\left(-2\right)^3+\left(3-\frac{9}{11}\right).1\frac{3}{8}\)

\(d)\left(-\frac{3}{4}\right)^2:\frac{-3}{8}+\frac{1}{2}-\frac{3}{4}-\left(\frac{-78}{57}\right)^0\)

2. tìm x

\(a)x+\frac{-1}{5}=\left(-\frac{3}{4^{ }}\right)^2\) \(b)\left|\frac{5}{2}x+\frac{2}{3}\right|-\frac{1}{4}=0\) \(c)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}+\frac{1}{2}x\) \(d)\left(x-\frac{1}{4}\right)^4=\frac{1}{81}\)

\(e)4x+3\frac{1}{4}=x-\frac{1}{4}\) \(g)\left(x-\frac{1}{3}\right)^3=\frac{1}{27}\)

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

Điều kiện: $ - \frac{1}{3} \le x \le 6$
Ta nhẩm thấy x = 5 là nghiệm của PT, thêm bớt và trục căn thức ta có:
Phương trình $ \Leftrightarrow \left( {\sqrt {3x + 1} - 4} \right) - \left( {\sqrt {6 - x} - 1} \right) + \left( {3{x^2} - 14x - 5} \right) = 0$

$ \Leftrightarrow \frac{{3\left( {x - 5} \right)}}{{\sqrt {3x + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)\left( {x - 5} \right) = 0$

$ \Leftrightarrow \left( {x - 5} \right)\left[ {\frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)} \right] = 0 \Leftrightarrow \left( {x - 5} \right)g\left( x \right) = 0$

Với điều kiện trên ta thấy g(x) > 0 vậy x = 5 là nghiệm của PT.