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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....
Đọc tiếp

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

12
20 tháng 3 2020

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)

PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)

Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)

giai tiep

20 tháng 3 2020

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)

NV
26 tháng 11 2019

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)

NV
26 tháng 11 2019

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)