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NV
16 tháng 8 2020

8.

ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)

\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)

\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)

\(\Leftrightarrow x=6\)

NV
16 tháng 8 2020

6.

ĐKXD: ...

\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)

\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)

\(\Leftrightarrow x=3\)

7.

\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)

\(\Rightarrow a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)

\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)

Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)

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:

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

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

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

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

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

NV
27 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

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

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

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

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

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

NV
27 tháng 10 2019

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

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
7 tháng 11 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)

\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))

\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(2\le x\le5\)

\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)

\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)

\(\Leftrightarrow...\)

NV
7 tháng 11 2019

c/ ĐKXĐ: \(x\le12\)

\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)

\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)

Xét (1):

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)

\(\Leftrightarrow a^3+a^2-12a=0\)

\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)

1 tháng 12 2019

TXĐ: D=R

\(\Leftrightarrow2x^2+6-2\sqrt{2x^2-3x+2}=3\left(x+4\right)\)

\(\Leftrightarrow\frac{2x^2-3x-6}{2}-4=\sqrt{2x^2-3x+2}-4\)

\(\Leftrightarrow\frac{2x^2-3x-14}{2}=\frac{2x^2-3x-14}{\sqrt{2x^2-3x+2}+4}\)

\(\left[{}\begin{matrix}2x^2-3x-14=0\\\frac{1}{2}=\frac{1}{\sqrt{2x^2-3x+2}+4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-2\\x=\frac{7}{2}\end{matrix}\right.\\\text{ pt vô nghiệm}\end{matrix}\right.\)

Vậy ....