1. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$

$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$

$\Leftrightarrow 22=10\sqrt{x-4}$

$\Leftrightarrow 2,2=\sqrt{x-4}$

$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$

(thỏa mãn)

2. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$

$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$

$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$

$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)

3. ĐKXĐ: $x\geq 3$

Bình phương 2 vế thu được:

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

$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$

$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$

$\Leftrightarrow (x-4)(7x+4)=0$

Do $x\geq 3$ nên $x=4$

Thử lại thấy thỏa mãn

Vậy $x=4$

4. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$

$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$

Ta thấy, với mọi $x\geq 4$ thì:

$(\sqrt{x}-2)^2\ge 0$

$2021\sqrt{x-4}\geq 0$ 

Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4+\left(4x-4\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\)

\(=\dfrac{x+4\sqrt{x}+4x\sqrt{x}+16x-4x-16\sqrt{x}}{x-16}\)

\(=\dfrac{13x+4x\sqrt{x}-12\sqrt{x}}{x-16}\)

Sửa: \(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)

\(C=\dfrac{\sqrt{x}+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\\ C=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)

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

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

\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)

\(=\dfrac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}}\)

=6

Ta có: \(\sqrt{x+4\sqrt{x}+4}+\sqrt{x-4\sqrt{x}+4}=4\)

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

\(\Leftrightarrow\left|\sqrt{x}-2\right|=2-\sqrt{x}\)

\(\Leftrightarrow0\le x< 4\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)