\(\Leftrightarrow30\sqrt{2}.\sqrt{x}=9\)

\(\Leftrightarrow\sqrt{x}=\dfrac{3\sqrt{2}}{20}\)

\(\Leftrightarrow x=\dfrac{9}{200}\)

Vậy pt có tập nghiệm là S=\(\left\{\dfrac{9}{200}\right\}\)

Lời giải:

a) ĐK: $x\geq 2$

PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$

\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)

Vậy..........

b) ĐK: $x\geq 0$

PT $\Leftrightarrow (\sqrt{x}-3)^2=0$

$\Leftrightarrow \sqrt{x}-3=0$

$\Leftrightarrow x=9$ (thỏa mãn)

c) ĐK: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$

$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$

$\Leftrightarrow 3\sqrt{x-3}=7$

$\Leftrightarrow x-3=(\frac{7}{3})^2$

$\Rightarrow x=\frac{76}{9}$

d)

ĐK: $x\geq \frac{-1}{2}$

PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$

$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$

$\Leftrightarrow 3\sqrt{2x+1}=6$

$\Leftrightarrow \sqrt{2x+1}=2$

$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)

cảm ơn nha <3

ai trả lời dùm em cái ak. E cảm ơn nhiều

a.\(\sqrt{x-2}=\sqrt{4-x}\)

đk: \(\left\{{}\begin{matrix}x-2\ge0\\4-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le4\end{matrix}\right.\Leftrightarrow2\le x\le4\)

pt đã cho tương đương với

\(x-2=4-x\)

\(\Leftrightarrow2x=6\Rightarrow x=3\left(TM\right)\)

b.\(\sqrt{x^2-8x+6}=x+2\)

đk: \(x+2\ge0\Rightarrow x\ge-2\)

pt đã cho tương đương với

\(x^2-8x+6=\left(x+2\right)^2\)

\(\Leftrightarrow x^2-8x+6=x^2+4x+4\)

\(\Leftrightarrow-12x=-2\Rightarrow x=\frac{1}{6}\left(TM\right)\)

c.\(\sqrt{2x-1}+5=\sqrt{8x-4}\)

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

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

\(\Leftrightarrow\sqrt{2x-1}=5\)

đk: \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

pt tương đương: \(2x-1=25\)

\(\Leftrightarrow2x=26\Rightarrow x=13\left(TM\right)\)

d.\(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)

\(\Leftrightarrow\sqrt{16\left(1-2x\right)}-\sqrt{4.3x}=\sqrt{3x}+\sqrt{9\left(1-2x\right)}\)

\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}+3\sqrt{1-2x}\)

\(\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}\)

đk: \(\left\{{}\begin{matrix}1-2x\ge0\\3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{1}{2}\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\frac{1}{2}\)

pt tương đương: \(1-2x=9.3x\)

\(\Leftrightarrow29x=1\Rightarrow x=\frac{1}{29}\left(TM\right)\)

e. \(\sqrt{x^2-9}-\sqrt{4x-12}=0\)

đk: \(\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ge0\\4x-12\ge0\end{matrix}\right.\Leftrightarrow x\ge3\)

pt đã cho tương đương với

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

\(\Leftrightarrow\sqrt{x-3}.\sqrt{x+3}-2\sqrt{x-3}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\left(TM\right)\\\sqrt{x+3}=2\Leftrightarrow x+3=4\Rightarrow x=1\left(KTM\right)\end{matrix}\right.\)

Nếu bạn tinh mắt một chút sẽ thấy:

Câu a: \(5\sqrt{2x-1}+2\sqrt{2x-1}-3\sqrt{x}=6\sqrt{2x-1}-2\sqrt{x}\)

Tương đương \(\sqrt{2x-1}=\sqrt{x}\Leftrightarrow\hept{\begin{cases}2x-1=x\\x\ge0\end{cases}}\Leftrightarrow x=1\).

Câu b: \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\).

Tương đương \(\sqrt{x-5}=\sqrt{1-x}\Leftrightarrow\hept{\begin{cases}x\le1\\x-5=1-x\end{cases}}\) (vô nghiệm)

Câu c: \(\sqrt{\left(x+3\right)\left(x-3\right)}-2\sqrt{x-3}=0\)

Tương đương \(\orbr{\begin{cases}x-3=0\\\sqrt{x+3}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Ấy chết! Sai ngu ở pt c rồi. Không có nghiệm \(x=1\) nha bạn.

a) ĐKXĐ : \(x\ge0\)

Ta có : \(\sqrt{3x}-\sqrt{27}+\sqrt{75x}=3\Leftrightarrow\sqrt{x}\left(\sqrt{3}+\sqrt{75}\right)=3+\sqrt{27}\)

\(\Leftrightarrow\sqrt{x}=\frac{3+\sqrt{27}}{\sqrt{3}+\sqrt{75}}=\frac{\sqrt{3}+3}{6}\)

\(\Leftrightarrow x=\frac{\left(3+\sqrt{3}\right)^2}{36}\)

b) ĐKXĐ : \(x\ge1\)

\(\sqrt{x-1}-\sqrt{4x-4}+\sqrt{9x-9}=10\)

\(\Leftrightarrow\sqrt{x-1}-\sqrt{4.\left(x-1\right)}+\sqrt{9.\left(x-1\right)}=10\)

\(\Leftrightarrow\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=10\)

\(\Leftrightarrow\sqrt{x-1}=5\Leftrightarrow x=26\) (TMĐK)

c) ĐKXĐ: \(x\ge-\frac{1}{2}\)

\(\sqrt{2x+1}+\sqrt{18x+9}-\sqrt{50x+25}=-3\)

\(\Leftrightarrow\sqrt{2x+1}+\sqrt{9\left(2x+1\right)}-\sqrt{25\left(2x+1\right)}=-3\)

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

\(\Leftrightarrow0=-3\) (Vô lí - loại)

Vậy pt vô nghiệm.

\(\sqrt{x-1}=5\)

\(\Leftrightarrow x-1=25\) (bình phương 2 vế)

\(\Leftrightarrow x=26\)

1) \(\sqrt{2-3x}+\sqrt{8-12x}=3\) (1) ĐKXĐ: \(x\le\dfrac{2}{3}\)

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

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

\(\Leftrightarrow3\sqrt{2-3x}=3\)

\(\Leftrightarrow\sqrt{2-3x}=1\)

\(\Leftrightarrow2-3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\) (Thỏa mãn)

Vậy \(x=\dfrac{1}{3}\) để \(\sqrt{2-3x}+\sqrt{8-12x}=3\)

2) \(4\sqrt{2x}+10\sqrt{8x}-9\sqrt{8x}+20=-10\) (2) ĐKXĐ: \(x\ge0\)

(2)\(\Leftrightarrow4\sqrt{2x}+20\sqrt{2x}-18\sqrt{2x}=-30\)

\(\Leftrightarrow6\sqrt{2x}=-30\)

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

Vì \(\sqrt{2x}\ge0\) với mọi x

\(\Rightarrow\) Không có giá trị của x để \(4\sqrt{2x}+10\sqrt{8x}-9\sqrt{8x}+20=-10\)

a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)

đk: x >/ 0

(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)

Kl: \(x=\dfrac{392}{169}\)

b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)

đk: x >/ 5

(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)

Kl: x=9

c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)

Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)

(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)

Kl: x=-6

d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)

Đk: \(x\ge\dfrac{4}{5}\)

(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)

Kl: x=12