\(a,\sqrt{\left(4x\right)^2}=3^2\)

=> \(\left|4x\right|=9\)

=> \(\left[{}\begin{matrix}4x=9\\4x=-9\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{9}{4}\\x=-\frac{9}{4}\end{matrix}\right.\)

\(b,\sqrt{4x^2}=x-1\)

=> \(\sqrt{\left(4x^2\right)^2}=x-1\)

=> \(\left|4x^2\right|=x-1\)

\(\)=> \(\left[{}\begin{matrix}4x^2=x-1\\4x^2=x+1\end{matrix}\right.\)

=> :))

c, :))

\(a.\)\(\left(\sqrt{4x^2}\right)^2=3^2\Leftrightarrow4x^2=9\Leftrightarrow x^2=\frac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy ...............

\(b.\left(\sqrt{4x^2}\right)^2=\left(x-1\right)^2\Leftrightarrow4x^2=x^2-2x+1\Leftrightarrow3x^2+2x-1=0\Leftrightarrow\left(x-\frac{1}{3}\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy ..............

a)\(2\sqrt{3}-\sqrt{4+x^2}=0\)

\(\Leftrightarrow\sqrt{12}-\sqrt{4+x^2}=0\)

\(\Leftrightarrow\sqrt{4+x^2}=\sqrt{12}\)

\(\Leftrightarrow4+x^2=12\Leftrightarrow x^2=8\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\)

vậy ....

b)\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18x}=0\) điều kiện xác định x\(\ge0\)

\(\Leftrightarrow3\sqrt{2x}+5\sqrt{4}\sqrt{2x}-\sqrt{9}\sqrt{2x}=20\)

\(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}=20\)

\(\Leftrightarrow10\sqrt{2x}=20\Leftrightarrow\sqrt{2x}=2\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\) (tm)

Vậy ....

c)\(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow4\left(x+2\right)^2=64\)

\(\Leftrightarrow\left(x+2\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy ...

a) pt <=> \(\sqrt{4+x^2}=2\sqrt{3}\)

<=> x² + 4 = 12

<=> x² = 8

<=> x = \(\pm2\sqrt{2}\)

b) ĐKXĐ: x ≥ 0

pt <=> \(3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}=20\)

<=> \(10\sqrt{2x}\) = 20

<=> \(\sqrt{2x}=2\)

<=> x = 2 (TM)

c) pt <=> 2|x + 2| = 8

<=> |x + 2| = 4

<=> \(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

d) ĐKXĐ: x ≥ 2

pt <=> \(\sqrt{x-2}=3\sqrt{x^2-4}\)

<=> 9x² - 12 = x - 2

<=> 9x² - x - 10 = 0

<=> 9(x + 1)(x - \(\dfrac{10}{9}\)) = 0

<=> \(\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{9}\end{matrix}\right.\)(KTM)

e) pt <=> 4x + 1 = -7

<=> 4x = -8

<=> x = -2

Lời giải:
ĐKXĐ:........

Bình phương 2 vế ta có:

\(\Rightarrow x+(9-x)+2\sqrt{x(9-x)}=-x^2+9x+9\)

\(\Leftrightarrow 2\sqrt{x(9-x)}=-x^2+9x=x(9-x)\)

\(\Leftrightarrow \sqrt{x(9-x)}(2-\sqrt{x(9-x)})=0\)

\(\Rightarrow \left[\begin{matrix} \sqrt{x}=0(1)\\ \sqrt{9-x}=0(2)\\ 2=\sqrt{x(9-x)}(3)\end{matrix}\right.\)

Với \((1)\Rightarrow x=0\) (t/m)

Với (2)\(\Rightarrow x=9\) (t/m)

Với (3): \(\Rightarrow 4=x(9-x)\)

\(\Leftrightarrow x^2-9x+4=0\)

\(x=\frac{9\pm \sqrt{65}}{2}\) (đều thỏa mãn)

Vậy............

\(m^3=4+\sqrt{80}-\sqrt{80}+4-3m\left(\sqrt[3]{4+\sqrt{80}}-\sqrt[3]{\sqrt{80}-4}\right)\)

\(\Leftrightarrow m^3=-12m+8\Leftrightarrow m^3+12m-8=0\)

vậy m la nghiệm của pt

ĐKXD: x, y > 0.

\(Pt_{\left(1\right)}\Leftrightarrow\dfrac{y+\sqrt{x}}{x}=\dfrac{2\left(y+\sqrt{x}\right)}{y}\Leftrightarrow\left(y+\sqrt{x}\right)\left(\dfrac{1}{x}-\dfrac{2}{y}\right)=0\)

\(\Rightarrow y=2x\), thế vào Pt(2): \(2x\left(\sqrt{x^2+1}-1\right)=\sqrt{3x^2+3}\)

Đặt \(\sqrt{x^2+1}=a\) thì \(\left\{{}\begin{matrix}2x\left(a-1\right)=\sqrt{3}a\\a^2-x^2=1\end{matrix}\right.\)

Giải ra ta được \(\left(a-2\right)\left(4a^3-3a+2\right)=0\) nhưng vì \(a\ge1\) nên a=2

Do đó \(x=\pm\sqrt{3}\). Vậy (x, y)=...

1) \(\sqrt{x^2+1}=\sqrt{5}\)

\(\Leftrightarrow x^2+1=5\)

\(\Leftrightarrow x^2=5-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2\)

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

2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\)) 

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=3+1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=\dfrac{4}{2}\)

\(\Leftrightarrow x=2\left(tm\right)\)

3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))

\(\Leftrightarrow43-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1=43-x\)

\(\Leftrightarrow x^2-x-42=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))

\(\Leftrightarrow\sqrt{4x-3}=x-2\)

\(\Leftrightarrow4x-3=\left(x-2\right)^2\)

\(\Leftrightarrow x^2-4x+4=4x-3\)

\(\Leftrightarrow x^2-8x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))

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

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1^2\)

\(\Leftrightarrow x=1\left(tm\right)\)

1)

\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy PT có nghiệm `x=2` hoặc `x=-2`

2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy PT có nghiệm `x=2`

3)

\(ĐKXĐ:x\le43\)

PT trở thành:

\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm `x=-6` hoặc `x=7`

4)

ĐKXĐ: \(x\ge\dfrac{3}{4}\)

PT trở thành:

\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)

5) 

ĐKXĐ: \(x\ge0\)

PT trở thành:

\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

Khi đó:

(1)\(\Leftrightarrow3t^2+8t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)

Vậy PT vô nghiệm.

Bài 1:

$\sqrt{x^2+1}=\sqrt{5}$

$\Leftrightarrow x^2+1=5$

$\Leftrightarrow x^2-4=0$

$\Leftrightarrow (x-2)(x+2)=0$

$\Leftrightarrow x-2=0$ hoặc $x+2=0$

$\Leftrightarrow x=\pm 2$ (đều tm)

2. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow 2x-1=3$

$\Leftrightarrow 2x=4$

$\Leftrightarrow x=2$ (tm) 

3. ĐKXĐ: $x\leq 43$

PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 43-x=(x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-42=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+6)(x-7)=0\end{matrix}\right.\)

$\Rightarrow x=7$ (tm) 

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)