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.

Mình rút gọn như thế này đúng không nhỉ?

\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)

\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)

\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)

Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)

1. \(x^3-x^2+12x\sqrt{x-1}+20=0\)

2. \(x^3+\sqrt{\left(x-1\right)^3}=9x+8\)

3. \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)

4. \(x^6+\left(x^3-3\right)^3=3x^5-9x^2-1\)

5. \(x^2-6\left(x+3\right)\sqrt{x+1}+14x+3\sqrt{x+1}+13=0\)

6. \(x^2-4x+\left(x-3\right)\sqrt{x^2-x+1}=-1\)

7. \(\sqrt{2x-1}+\sqrt{5-x}=x-2+2\sqrt{-2x^2+11x-5}\)

8. \(\sqrt{5x+11}-\sqrt{6-x}+5x^2-14x-60=0\)

9. \(x^2+6x+8=3\sqrt{x+2}\)

10. \(2x^2+3x-2=\left(2x-1\right)\sqrt{2x^2+x-3}\)

11. \(\sqrt{x+1}+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=1\)

12. \(x^2-\sqrt{x^2-4x}=4\left(x+3\right)\)

13. \(x^2-x-4=2\sqrt{x-1}\left(1-x\right)\)

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\)

15. \(\sqrt{2x^2+3x+2}+\sqrt{4x^2+6x+21}=11\)

16. \(\sqrt{x+3+3\sqrt{2x-3}}+\sqrt{x-1+\sqrt{2x-1}}=2\sqrt{2}\)

17. \(\left(x-2\right)^2\left(x-1\right)\left(x-3\right)=12\)

18. \(2x^2+\sqrt{x^2-2x-19}=4x+74\)

19. \(x^4+x^2-20=0\)

20. \(x+\sqrt{4-x^2}=2+3x\sqrt{4-x^2}\)

21. \(\left(x^2+x+1\right)\left(\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1\right)=9\)

22. \(\sqrt{x^2-3x+5}+x^2=3x+7\)

23. \(x^2+6x+5=\sqrt{x+7}\)

24. \(\frac{2x^2-3x+10}{x+2}=3\sqrt{\frac{x^2-2x+4}{x+2}}\)

25. \(5\sqrt{x-1}-\sqrt{x+7}=3x-4\)

26. \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)

27. \(\sqrt{x-1}+\sqrt{5-x}-2=2\sqrt{\left(x-1\right)\left(5-x\right)}\)

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\)

29. \(\frac{26x+5}{\sqrt{x^2+30}}+2\sqrt{26x+5}=3\sqrt{x^2+30}\)

30. \(\frac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\frac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)

b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)

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

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

\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)