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Bài 1:
ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
Bài 2:
a. ĐKXĐ: $x\geq \frac{1}{3}$
PT $\Leftrightarrow 3x-1=2^2=4$
$\Leftrightarrow x=\frac{5}{3}$ (tm)
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}+2\sqrt{x-2}=6$
$\Leftrightarrow 3\sqrt{x-2}=6$
$\Leftrightarrow \sqrt{x-2}=2$
$\Leftrightarrow x-2=4$
$\Leftrightarrow x=6$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$
PT $\Leftrightarrow 5x-2=7^2=49$
$\Leftrightarrow 5x=51$
$\Leftrightarrow x=\frac{51}{5}=10,2$
b. ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$
$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$
$\Leftrightarrow 8\sqrt{x-3}=24$
$\Leftrightarrow \sqrt{x-3}=3$
$\Leftrightarrow x-3=9$
$\Leftrightarrow x=12$ (tm)
Bài 1:
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$
$\Leftrightarrow (x-2)(x-3)-2.\frac{x-3}{\sqrt{x-2}+1}=0$
$\Leftrightarrow (x-3)[(x-2)-\frac{2}{\sqrt{x-2}+1}]=0$
$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$
Nếu $x-3=0$
$\Leftrightarrow x=3$ (tm)
Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$
$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)
$\Leftrightarrow a^3+a^2-2=0$
$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$
$\Leftrightarrow (a-1)(a^2+2a+2)=0$
Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$
$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)
Vậy pt có nghiệm duy nhất $x=3$.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Đặt:z=\dfrac{1}{\sqrt{y}-3}\left(y\ge0;y\ne9\right)\\ \left\{{}\begin{matrix}x+2+\dfrac{2}{\sqrt{y}-3}=9\\2x+4-\dfrac{1}{\sqrt{y-3}}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2z=9-2=7\\2x-z=8-4=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+4z=14\\2x-z=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5z=10\\2x-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=2\\x=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{y}-3}=2\\x=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\sqrt{y}-6=1\\x=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=\dfrac{7}{2}\\x=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\left(\dfrac{7}{2}\right)^2=\dfrac{49}{4}\\x=3\end{matrix}\right.\)
Anh giải hệ lun hi, chứ ĐKXĐ là: \(\left(y\ge0;y\ne9\right)\)
\(ĐKXĐ: \begin{cases} \sqrt{y}-3 \ne 0\\\sqrt{y}\ge0\end{cases} \Leftrightarrow \begin{cases} y\ne9\\y\ge0 \end{cases}\)
1) \(\sqrt{x+3}=2x+1\)(đk: \(x\ge-3\))
\(\Rightarrow x+3=\left(2x+1\right)^2\)
\(\Leftrightarrow4x^2+3x-2=0\)(1)
\(\Delta=3^2+3.2.4=41>0\)
Do đó (1) có hai nghiệm phân biệt \(x_{1,2}=\frac{-3\pm\sqrt{41}}{8}\)
Thử lại chỉ có \(x=\frac{-3+\sqrt{41}}{8}\)thỏa mãn.
2) \(\sqrt{x+2}=x\)(đk: \(x\ge-2\))
\(\Rightarrow x+2=x^2\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Thử lại chỉ có \(x=2\)thỏa mãn.