1. ĐKXĐ: $\frac{-1}{2}\leq x\leq 5$
PT \(\Leftrightarrow [(2x+1)-6\sqrt{2x+1}+9]+[(5-x)-2\sqrt{5-x}+1]=0\)
\(\Leftrightarrow (\sqrt{2x+1}-3)^2+(\sqrt{5-x}-1)^2=0\)
Thấy rằng:
$(\sqrt{2x+1}-3)^2\geq 0$
$(\sqrt{5-x}-1)^2\geq 0$
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{2x+1}-3=\sqrt{5-x}-1=0$
$\Leftrightarrow x=4$ (tm)
2. ĐKXĐ: $x\geq 8$
PT \(\Leftrightarrow 2x-2\sqrt{x-8}-6\sqrt{x}+2=0\)
\(\Leftrightarrow (x-8)-2\sqrt{x-8}+1+[x-6\sqrt{x}+9]=0\)
\(\Leftrightarrow (\sqrt{x-8}-1)^2+(\sqrt{x}-3)^2=0\)
\(\Rightarrow \sqrt{x-8}-1=\sqrt{x}-3=0\Leftrightarrow x=9\) (thỏa mãn)
Vậy $x=9$
3. ĐKXĐ: $x\geq \frac{-3}{2}$
PT \(\Leftrightarrow (x^2-6x+9)-[(2x+3)-6\sqrt{2x+3}+9]=0\)
\(\Leftrightarrow (x-3)^2-(\sqrt{2x+3}-3)^2=0\)
\(\Leftrightarrow (x-\sqrt{2x+3})(x+\sqrt{2x+3}-6)=0\)
Nếu $x-\sqrt{2x+3}=0$
\(\Leftrightarrow \left\{\begin{matrix} x=\sqrt{2x+3}\geq 0\\ x^2=2x+3\end{matrix}\right.\Leftrightarrow x=3\)
Nếu $x+\sqrt{2x+3}-6=0$
\(\Leftrightarrow \left\{\begin{matrix} 6-x=\sqrt{2x+3}\geq 0\\ (6-x)^2=2x+3\end{matrix}\right.\Leftrightarrow x=3\)
Vậy $x=3$
4. ĐKXĐ: $x\geq -1$
Đặt $\sqrt{x+1}=a; \sqrt{x+2}=b(a,b\geq 0)$
PT trở thành:
$7a^2-5b^2+2ab=a+b$
$\Leftrightarrow (a+b)(7a-5b)=a+b$
$\Leftrightarrow (a+b)(7a-5b-1)=0$
Nếu $a+b=0$
$\Leftrightarrow \sqrt{x+1}+\sqrt{x+2}=0$
$\Leftrightarrow \sqrt{x+1}=-\sqrt{x+2}$
$\Rightarrow x+1=x+2$ (vô lý)
Nếu $7a-5b-1=0$
$\Leftfrightarrow 7\sqrt{x+1}=5\sqrt{x+2}+1$
$\Rightarrow 49(x+1)=25(x+2)+1+10\sqrt{x+2}$
$\Leftrightarrow 12x-1=5\sqrt{x+2}$
$\Rightarrow (12x-1)^2=25(x+2)$
$\Leftrightarrow x=\frac{7}{9}$ hoặc $x=\frac{-7}{16}$
Thử lại thấy $x=\frac{7}{9}$
5. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{x-1}-1+\sqrt{3x-2}-2=x^2-4\)
\(\Leftrightarrow \frac{x-2}{\sqrt{x-1}+1}+\frac{3(x-2)}{\sqrt{3x-2}+2}=(x-2)(x+2)\)
\(\Leftrightarrow (x-2)(x+2-\frac{3}{\sqrt{3x-2}+2}-\frac{1}{\sqrt{x-1}+1})=0\)
Ta thấy:
\(\frac{3}{\sqrt{3x-2}+2}+\frac{1}{\sqrt{x-1}+1}\leq \frac{3}{2}+\frac{1}{1}=\frac{5}{2}< 3\leq x+2\) nên biểu thức \(x+2-\frac{3}{\sqrt{3x-2}+2}-\frac{1}{\sqrt{x-1}+1}>0\)
$\Rightarrow x-2=0$
$\Leftrightarrow x=2$ (tm)
6. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow 2x^2+5x-1=7\sqrt{(x-1)(x^2+x+1)}\)
Đặt $\sqrt{x-1}=a; \sqrt{x^2+x+1}=b(a,b\geq 0)$
Khi đó pt trở thành:
$2b^2+3a^2=7ab$
$\Leftrightarrow 2b^2-7ab+3a^2=0$
$\Leftrightarrow (3a-b)(a-2b)=0$
$\Leftrightarrow 3a=b$ hoặc $a=2b$
Nếu $3a=b$
$\Leftrightarrow 9a^2=b^2$
$\Leftrightarrow 9(x-1)=x^2+x+1$
$\Leftrightarrow x=4\pm \sqrt{6}$ (tm)
Nếu $a=2b$
$\Leftrightarrow a^2=4b^2$
$\Leftrightarrow x-1=4(x^2+x+1)$
$\Leftrightarrow 4x^2+3x+5=0$ (pt này vô nghiệm)
Vậy..........
7. Nghiệm khá xấu, không phù hợp với tổng thể bài. Bạn xem lại nhé.
8. ĐKXĐ: $x\geq \frac{1}{2}$
PT \(\Leftrightarrow 2(\sqrt{x+3}-2\sqrt{2x-1})=x-1\)
\(\Leftrightarrow 2.\frac{x+3-4(2x-1)}{\sqrt{x+3}+2\sqrt{2x-1}}=x-1\)
\(\Leftrightarrow \frac{14(1-x)}{\sqrt{x+3}+2\sqrt{2x-1}}=x-1\)
\(\Leftrightarrow (x-1)\left(1+\frac{14}{\sqrt{x+3}+2\sqrt{2x-1}}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn lớn hơn $0$
$\Rightarrow x-1=0$
$\Leftrightarrow x=1$ (tm)
9. ĐKXĐ: $x\geq 2$
Đặt $\sqrt{x-2}=a; \sqrt{x-1}=b$ với $a,b\geq 0$
PT trở thành:
$a+2b=ab+2$
$\Leftrightarrow ab-a-2b+2=0$
$\Leftrightarrow (a-2)(b-1)=0$
$\Rightarrow a=2$ hoặc $b=1$
Nếu $a=2\Leftrightarrow x-2=4\Leftrightarrow x=6$ (tm)
Nếu $b=1\Leftrightarrow x-1=1\Leftrightarrow x=2$ (tm)
10. ĐKXĐ: $x\geq \frac{1}{2}$
PT \(\Leftrightarrow (\sqrt{2x-1}-1)+(\sqrt{3x+1}-2)=4(x-1)\)
\(\Leftrightarrow \frac{2(x-1)}{\sqrt{2x-1}+1}+\frac{3(x-1)}{\sqrt{3x+1}+2}=4(x-1)\)
\(\Leftrightarrow (x-1)\left[\frac{2}{\sqrt{2x-1}+1}+\frac{3}{\sqrt{3x+1}+2}-4\right]=0\)
Dễ thấy \(\frac{2}{\sqrt{2x-1}+1}+\frac{3}{\sqrt{3x+1}+2}-4\leq \frac{2}{1}+\frac{3}{2}-4< 0\)
$\Rightarrow x-1=0$
$\Leftrightarrow x=1$ (tm)
11. ĐKXĐ: $x\geq \frac{-1}{2}$
PT \(\Leftrightarrow x^2-[(2x+1)+2\sqrt{2x+1}+1]=0\)
\(\Leftrightarrow x^2-(\sqrt{2x+1}+1)^2=0\)
\(\Leftrightarrow (x-\sqrt{2x+1}-1)(x+\sqrt{2x+1}+1)=0\)
Với mọi $x\geq \frac{-1}{2}$ thì dễ thấy $x+\sqrt{2x+1}+1>0$
Do đó:
$x-\sqrt{2x+1}-1=0$
$\Lefrightarrow x-1=\sqrt{2x+1}$
\(\Leftrightarrow \left\{\begin{matrix}
x-1\geq 0\\
(x-1)^2=2x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\geq 1\\
x^2-4x=0\end{matrix}\right.\Leftrightarrow x=4\)
Vậy $x=4$
12. ĐKXĐ: $x\geq -3$
PT \(\Leftrightarrow (x+3)+2\sqrt{x+3}+1=4x^2+3x-3+(x+3)+1\)
\(\Leftrightarrow (\sqrt{x+3}+1)^2=4x^2+4x+1=(2x+1)^2\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x+3}+1=2x+1\\ \sqrt{x+3}+1=-(2x+1)\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \sqrt{x+3}=2x\\ \sqrt{x+3}+2x+2=0\end{matrix}\right.\)
Nếu $\sqrt{x+3}=2x$
$\Leftrightarrow$\(\left\{\begin{matrix} 2x\geq 0\\ x+3=4x^2\end{matrix}\right.\Leftrightarrow x=1\) (tm)
Nếu $\sqrt{x+3}+2x+2=0$
\(\Leftrightarrow \left\{\begin{matrix} 2x+2=-\sqrt{x+3}\leq 0\\ (2x+2)^2=x+3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq -1\\ 4x^2+7x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{-7-\sqrt{33}}{8}\)
13. ĐKXĐ: $x\geq \frac{-1}{8}$
PT \(\Leftrightarrow [(8x+1)-6\sqrt{8x+1}+9]+[(x+3)-4\sqrt{x+3}+4]=0\)
\(\Leftrightarrow (\sqrt{8x+1}-3)^2+(\sqrt{x+3}-2)^2=0\)
\(\Rightarrow \sqrt{8x+1}-3=\sqrt{x+3}-2=0\)
\(\Leftrightarrow x=1\) (tm)
14. ĐKXĐ: $x\geq -1$
Đặt $\sqrt{x+1}=a; \sqrt{x+2}=b; \sqrt{x-1}=c$ với $a,b,c\geq 0$ thì pt trở thành:
\(ab+ac+6=3a+2b+2c\)
\(\Leftrightarrow b(a-2)+c(a-2)-3(a-2)=0\)
\(\Leftrightarrow (b+c-3)(a-2)=0\)
Nếu $a-2=0$
$\Leftrightarrow x+1=4\Leftrightarrow x=3$ (tm)
Nếu $b+c-3=0$
$\Leftrightarrow \sqrt{x+2}+\sqrt{x-1}-3=0$
$\Leftrightarrow (\sqrt{x+2}-2)+(\sqrt{x-1}-1)=0$
$\Leftrightarrow \frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{x-1}+1}=0$
$\Leftrightarrow (x-2)(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{x-1}+1})=0$
$\Leftrightarrow x-2=0$ (do nhân tử thứ 2 dương)
$\Leftrightarrow x=2$ (tm)
15. ĐKXĐ: $1\leq x\leq 4$
Đặt $\sqrt{4-x}=a; \sqrt{2x-2}=b$ với $a,b\geq 0$ thì $x+5=2a^2+\frac{3}{2}b^2$ và $2a^2+b^2=6$.
PT trở thành:
$2a^2+\frac{3}{2}b^2+2ab=4(a+b)$
$\Leftrightarrow 4a^2+3b^2+4ab=8(a+b)$
$\Leftrightarrow 2(a+b)^2+2a^2+b^2=8(a+b)$
$\Leftrightarrow 2(a+b)^2+6=8(a+b)$
$\Leftrightarrow (a+b)^2+3-4(a+b)=0$
$\Leftrightarrow (a+b-1)(a+b-3)=0$
Nếu $a+b-1=0$
$\Leftrightarrow \sqrt{4-x}+\sqrt{2x-2}=1$
$\Rightarrow 2+x+2\sqrt{(4-x)(2x-2)}=1$
$\Leftrightarrow (x+1)+2\sqrt{(4-x)(2x-2)}=0$
PT này vô nghiệm do với $1\leq x\leq 4$ thì vế trái dương
Nếu $a+b-3=0$
$\Leftrightarrow \sqrt{4-x}+\sqrt{2x-2}-3=0$
$\Leftrightarrow (\sqrt{4-x}-1)+(\sqrt{2x-2}-2)=0$
$\Leftrightarrow \frac{3-x}{\sqrt{4-x}+1}+\frac{2(x-3)}{\sqrt{2x-2}+2}=0$
$\Leftrightarrow (x-3)(\frac{2}{\sqrt{2x-2}+2}-\frac{1}{\sqrt{4-x}+1})=0$
Nếu $x-3=0\Leftrightarrow x=3$ (tm)
Nếu \(\frac{2}{\sqrt{2x-2}+2}-\frac{1}{\sqrt{4-x}+1}=0\)
\(\Leftrightarrow 2\sqrt{4-x}=\sqrt{2x-2}\)
\(\Leftrightarrow 2(4-x)=x-1\)
$\Leftrightarrow x=3$ (tm)
Vậy.........
16.
ĐKXĐ: $x\geq 4$
PT \(\Leftrightarrow (\sqrt{x-4}-1)+(\sqrt[3]{x+3}-2)=0\)
\(\Leftrightarrow \frac{x-5}{\sqrt{x-4}+1}+\frac{x-5}{\sqrt[3]{(x+3)^2}+2\sqrt[3]{x+3}+4}=0\)
\(\Leftrightarrow (x-5)\left[\frac{1}{\sqrt{x-4}+1}+\frac{1}{\sqrt[3]{(x+3)^2}+2\sqrt[3]{x+3}+4}\right]=0\)
Dễ thấy biểu thức trong ngoặc lớn lớn hơn $0$
$\Rightarrow x-5=0$
$\Leftrightarrow x=5$
17. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow x^2+x-6=2(1-x)(\sqrt{x-1}-1)\)
\(\Leftrightarrow (x+3)(x-2)=2(1-x).\frac{x-2}{\sqrt{x-1}+1}\)
\(\Leftrightarrow (x-2)\left[(x+3)-\frac{2(1-x)}{\sqrt{x-1}+1}\right]=0\)
Với $x\geq 1$ thì:
\(x+3\geq 4; \frac{2(1-x)}{\sqrt{x-1}+1}\leq 0\) nên biểu thức trong ngoặc lớn luôn dương
$\Rightarrow x-2=0$
$\Leftrightarrow x=2$ (tm)
18.
ĐKXĐ $x\geq -1$
Đặt $x-2=a; \sqrt{x+1}=b$ thì pt trở thành:
$\sqrt{a^2+8b^2}=2a+b$
$\Rightarrow a^2+8b^2=4a^2+b^2+4ab$
$\Leftrightarrow 3a^2+4ab-7b^2=0$
$\Leftrightarrow (3a+7b)(a-b)=0$
Nếu $a-b=0\Leftrightarrow x-2=\sqrt{x+1}$
$\Rightarrow x-2\geq 0$ và $(x-2)^2=x+1$
$\Rightarrow x=\frac{5+\sqrt{13}}{2}$
Nếu $3a+7b=0$
Vì $2a+b=\sqrt{a^2+8b^2}> 0$ và $b\geq 0$ nên:
$3a+7b=1,5(2a+b)+5,5b>0$ nên th này vô nghiệm
Vậy.......
19. ĐKXĐ: $x\geq 2$
Đặt $\sqrt{x+6}=a; \sqrt{x-2}=b$ với $a,b\geq 0$ thì:
$(a-b)(1+ab)=a^2-b^2$
$\Leftrightarrow (a-b)[(1+ab)-(a+b)]=0$
$\Leftrightarrow (a-b)(a-1)(b-1)=0$
Nếu $a-b=0\Leftrightarrow x+6=x-2$ (vô lý)
Nếu $a-1=0\Leftrightarrow x+6=1\Leftrightarrow x=-5$ (không tm đkxđ)
Nếu $b-1=0\Leftrightarrow x-2=1\Leftrightarrow x=3$ (tm)
20. ĐK: $x\geq 0$ hoặc $x\leq -1$
PT $\Leftrightarrow 3(x^2+x)-2\sqrt{x^2+x}-1=0$
Đặt $\sqrt{x^2+x}=a$ với $a\geq 0$ thì pt trở thành:
$3a^2-2a-1=0$
$\Leftrightarrow (a-1)(3a+1)=0$
Nếu $a-1=0\Leftrightarrow x^2+x-1=0$
$\Leftrightarrow x=\frac{-1\pm \sqrt{5}}{2}$ (tm)
Nếu $3a+1=0\Leftrightarrow a=\frac{-1}{3}< 0$ (vô lý do $a\geq 0$)
Vậy.........
