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. \(\frac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\frac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)
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
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
Bạn ơi lần sau bạn đăng bài thì cố gắng đăng giãn giãn bớt bớt/ chia nhỏ bài ra chứ một cục bài như thế này nhìn rất đáng sợ và gây tâm lý ngại đọc nhé.
1. ĐKXĐ: $x\geq 1$
Đặt $x\sqrt{x-1}=a\Rightarrow x^3-x^2=x^2(x-1)=a^2$. PT đã cho trở thành:
$a^2+12a+20=0(*)$
Lại thấy rằng vì $x\geq 1$ nên $a\geq 0$
$\Rightarrow a^2+12a+20\geq 20>0$. Do đó $(*)$ vô nghiệm. Kéo theo PT ban đầu vô nghiệm.
2. ĐK: $x\geq -1$
$x^3+\sqrt{(x+1)^3}=9x+8$
$\Leftrightarrow x^3-9x-8+\sqrt{(x+1)^3}=0$
$\Leftrightarrow (x^2-x-8)(x+1)+(x+1)\sqrt{x+1}=0$
$\Leftrightarrow (x+1)(x^2-x-8+\sqrt{x+1})=0$
Nếu $x+1=0\Rightarrow x=-1$ (thỏa mãn)
Nếu $x^2-x-8+\sqrt{x+1}=0$
$\Leftrightarrow (x^2-9)-(x-3)+(\sqrt{x+1}-2)=0$
$\Leftrightarrow (x-3)\left(x+3+\frac{1}{\sqrt{x+1}+2}}-1\right)=0$
Dễ thấy với $x\geq -1$ thì biểu thức trong ngoặc lớn luôn lớn hơn $0$
Do đó $x-3=0\Rightarrow x=3$
Vậy $x=-1$ hoặc $x=3$
3.
ĐK: $x\in\mathbb{R}$
Từ PT dễ suy ra $3x>0\Rightarrow x>0$
PT $\Leftrightarrow \sqrt{2x^2+x+1}-2x+\sqrt{x^2-x+1}-x=0$
$\Leftrightarrow \frac{-2x^2+x+1}{\sqrt{2x^2+x+1}+2x}+\frac{-x+1}{\sqrt{x^2-x+1}+x}=0$
$\Leftrightarrow \frac{(1-x)(2x+1)}{\sqrt{2x^2+x+1}+2x}+\frac{(1-x)}{\sqrt{x^2-x+1}+x}=0$
$\Leftrightarrow (1-x)\left(\frac{2x+1}{\sqrt{2x^2+x+1}+2x}+\frac{1}{\sqrt{x^2-x+1}+x}\right)=0$
Với $x>0$ dễ thấy biểu thức trong ngoặc luôn lớn hơn $0$
Do đó $1-x=0\Rightarrow x=1$ (thỏa mãn)
Vậy.........
26. \(2\left(x^2+2\right)=5\sqrt{x^3+1}\) DK: \(x\ge-1\)
PT\(\Leftrightarrow2\left(x+1+x^2-x+1\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Dat \(\sqrt{x+1}=a\)(\(a\ge0\))
\(\sqrt{x^2-x+1}=b\left(b>0\right)\)
PTTT: \(2\left(a^2+b^2\right)=5ab\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\)
+) \(2a=b\) \(\Leftrightarrow2\sqrt{x+1}=\sqrt{x^2-x+1}\)
\(\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{37}}{2}\\x=\frac{5-\sqrt{37}}{2}\end{matrix}\right.\)
+) \(a=2b\Leftrightarrow\sqrt{x+1}=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow x+1=4x^2-4x+4\Leftrightarrow4x^2-5x+3=0\)(vo nghiem)
Vay.........
22.\(\sqrt{x^2-3x+5}+x^2=3x+7\)(DK: x∈R)
Dat \(\sqrt{x^2-3x+5}=a\left(a>0\right)\)
PTTT: \(a^2+a-12=0\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\left(loai\right)\end{matrix}\right.\)
+) \(a=3\Leftrightarrow\sqrt{x^2-3x+5}=3\Leftrightarrow x^2-3x-4=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vay S = \(\left\{-1;4\right\}\)
19. Bai nay ez ma
\(x^4+x^2-20=0\Leftrightarrow\left(x^2-4\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow x^2-4=0\)(\(x^2+5>0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
4.
Đặt $x^3-3=m, x^2=n$ thì PT trở thành:
$n^3+m^3=3n(m+3)-9n-1$
$\Leftrightarrow n^3+m^3+1-3mn=0$
$\Leftrightarrow (m+n+1)(m^2+n^2+1-mn-m-n)=0$ (đây là hằng đẳng thức quen thuộc)
Từ đây ta xét 2TH:
TH1: $m+n+1=0$
$\Leftrightarrow x^3-3+x^2+1=0$
$\Leftrightarrow x^3+x^2-2=0$
$\Leftrightarrow x^2(x-1)+2(x-1)(x+1)=0$
$\Leftrightarrow (x-1)(x^2+2x+2)=0\Rightarrow x=1$ (tm)
TH2: $m^2+n^2+1-mn-m-n=0$
$\Leftrightarrow (m-1)^2+(n-1)^2+(m-n)^2=0$
$\Rightarrow m=n=1$. Thử vào thấy vô lý nên loại
Vậy.........
5.
ĐK: $x\geq -1$
PT $\Leftrightarrow (x^2+14x+13)-6(x+3)\sqrt{x+1}+3\sqrt{x+1}=0$
$\Leftrightarrow (x+1)(x+13)-6(x+3)\sqrt{x+1}+3\sqrt{x+1}=0$
$\Leftrightarrow \sqrt{x+1}[(x+13)\sqrt{x+1}-6(x+3)+3]=0$
$\Rightarrow \sqrt{x+1}=0$ hoặc $(x+13)\sqrt{x+1}-6(x+3)+3=0$
Nếu $\sqrt{x+1}=0\Rightarrow x=-1$ (tm)
Nếu $(x+13)\sqrt{x+1}-6(x+3)+3=0$
$\Leftrightarrow (x+13)(\sqrt{x+1}-3)-3(x-8)=0$
$\Leftrightarrow (x+13).\frac{x-8}{\sqrt{x+1}+3}-3(x-8)=0$
$\Leftrightarrow (x-8)\left(\frac{x+13}{\sqrt{x+1}+3}-3\right)=0$
$\Leftrightarrow (x-8).\frac{x+4-3\sqrt{x+1}}{\sqrt{x+1}+3}=0$
$\Rightarrow x-8=0$ hoặc $x+4-3\sqrt{x+1}=0$
Nếu $x-8=0\Rightarrow x=8$ (tm)
Nếu $x+4-3\sqrt{x+1}=0\Leftrightarrow (x+1)-3\sqrt{x+1}+3=0$
$\Leftrightarrow (\sqrt{x+1}-\frac{3}{2})^2=-\frac{3}{4}< 0$ (vô lý)
Vậy $x=-1$ hoặc $x=8$
9. \(x^2+6x+8=3\sqrt{x+2}\)(DK: \(x\ge-2\))
PT\(\Leftrightarrow x^2+6x+5=3\left(\sqrt{x+2}-1\right)\Leftrightarrow\left(x+1\right)\left(x+5\right)=3.\frac{x+1}{\sqrt{x+2}+1}\)
\(\Leftrightarrow\left(x+1\right)\left(x+5-\frac{3}{\sqrt{x+2}+1}\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x+5=\frac{3}{\sqrt{x+2}+1}\left(1\right)\end{matrix}\right.\)
(1) co VT\(\ge5-2=3\)
VP\(\le\frac{3}{1}=3\)
Suy ra VT=VP=3\(\Leftrightarrow x=-2\)(tm)
Vay S = \(\left\{-1;-2\right\}\)
11. \(\sqrt{x+1}+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=1\) (DK: \(-1\le x\le4\))
PT\(\Leftrightarrow\left(\sqrt{4-x}-1\right)\left(1-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4-x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\left(TM\right)\)
Dài chưa chắc đã hay. Đề nghị xóa khỏi câu hỏi hay :D
