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. \(\frac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\frac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)
Tiếp =))
c)Áp dụng BĐT AM-GM ta có:
\(x\sqrt{y-1}\le\frac{x\left(y-1+1\right)}{2}=\frac{xy}{2}\)
\(2y\sqrt{x-1}\le\frac{2y\left(x-1+1\right)}{2}=\frac{2xy}{2}\)
Cộng theo vế 2 BĐT trên ta có:
\(VT=x\sqrt{y-1}+2y\sqrt{x-1}\le\frac{3xy}{2}=VP\)
Nên xảy ra khi \(x=y\) thay vào giải ra có: x=y=2
d)\(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)
\(pt\Leftrightarrow\sqrt{2x^2+x+1}-2+\sqrt{x^2-x+1}-1=3x-3\)
\(\Leftrightarrow\frac{2x^2+x+1-4}{\sqrt{2x^2+x+1}+2}+\frac{x^2-x+1-1}{\sqrt{x^2-x+1}+1}=3\left(x-1\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x\left(x-1\right)}{\sqrt{x^2-x+1}+1}-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x}{\sqrt{x^2-x+1}+1}-3\right)=0\)
pt trong ngoặc vn nên x=1
Tắm đã làm nốt cho :))
Chả ai giúp t gank =)), mà lần sau đăng ít 1 thôi đăng lắm thế này nhìn nản cmn luôn ấy
a)\(\sqrt{x^2+x-5}+\sqrt{-x^2+x+3}=x^2-3x+4\)
\(pt\Leftrightarrow\sqrt{x^2+x-5}-1+\sqrt{-x^2+x+3}-1=x^2-3x+2\)
\(\Leftrightarrow\frac{x^2+x-5-1}{\sqrt{x^2+x-5}+1}+\frac{-x^2+x+3-1}{\sqrt{-x^2+x+3}+1}=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+3\right)}{\sqrt{x^2+x-5}+1}+\frac{-\left(x-2\right)\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\frac{\left(x+3\right)}{\sqrt{x^2+x-5}+1}-\frac{\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\right]=0\)
Pt trong ngoặc <0 nên x=2 là nghiệm
b)\(\frac{x^2}{2}+\frac{x}{2}+1=\sqrt{2x^3-x^2+x+1}\)\
Đk:\(x\ge-\frac{1}{2}\)
\(\Leftrightarrow\frac{x^2}{2}+\frac{x}{2}+1-\left(2x+1\right)=\sqrt{2x^3-x^2+x+1}-\left(2x+1\right)\)
\(\Leftrightarrow\frac{x^2}{2}+\frac{x}{2}+1-\left(2x+1\right)=\frac{2x^3-x^2+x+1-\left(2x+1\right)^2}{\sqrt{2x^3-x^2+x+1}+2x+1}\)
\(\Leftrightarrow\frac{x^2-3x}{2}-\frac{2x^3-5x^2-3x}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{2}-\frac{x\left(x-3\right)\left(2x+1\right)}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)
\(\Leftrightarrow x\left(x-3\right)\left(\frac{1}{2}-\frac{2x+1}{\sqrt{2x^3-x^2+x+1}+2x+1}\right)=0\)
Pt trong ngoặc vô nghiệm nốt nên
\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)