\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)

\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)

b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)

a: \(A=\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{4+\sqrt{3}}{5-2\sqrt{3}}}\)

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

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)

b: \(B=\dfrac{x\sqrt{x}-2x+28}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}-\dfrac{x-16}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4\sqrt{x}-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...

Lời giải:

a. ĐKXĐ: $x\geq 0$

$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$

$\Leftrightarrow 13\sqrt{2x}=28$

$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$

$\Leftrightarrow 2x=\frac{784}{169}$

$\Leftrightarrow x=\frac{392}{169}$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$

PT $\Leftrightarrow \frac{3x-2}{x+1}=9$

$\Rightarrow 3x-2=9(x+1)$

$\Leftrightarrow x=\frac{-11}{6}$ (tm)

b: \(\Leftrightarrow\left(x^2+5x+4\right)=5\sqrt{x^2+5x+28}\)

Đặt \(x^2+5x+4=a\) 

Theo đề, ta có \(5\sqrt{a+24}=a\)

=>25a+600=a²

=>a=40 hoặc a=-15

=>x²+5x-36=0

=>(x+9)(x-4)=0

=>x=4 hoặc x=-9

c: \(\Leftrightarrow x^2+5x=2\sqrt[3]{x^2+5x-2}-2\)

Đặt \(x^2+5x=a\)

Theo đề, ta có: \(a=2\sqrt[3]{a}-2\)

\(\Leftrightarrow\sqrt[3]{8a}=a+2\)

=>(a+2)³=8a

=>\(a^3+6a^2+12a+8-8a=0\)

\(\Leftrightarrow a^3+6a^2+4a+8=0\)

Đến đây thì bạn chỉ cần bấm máy là xong

a) ĐKXĐ: \(3\le x\le10\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-4\\x\ne4\end{matrix}\right.\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ne4\end{matrix}\right.\)

d) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

e) ĐKXĐ: \(x\in R\)

a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)

b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)

c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)