b, \(\sqrt{x^{2^{ }}-5x-14}\ge2x-1\)

*TH1:

+, \(x^{2^{ }}-5x-14\ge0\)

+, \(2x-1< 0\)

*TH2:

+, \(2x-1\ge0\)

+, \(x^2-5x-14\ge\left(2x-1\right)^2\)

Câu b bạn giải theo 2 trường hợp này là được nhé

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

\(\Leftrightarrow2x^2+x-3+2x-\sqrt{5x-1}+\sqrt[3]{x-9}+2\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{4x^2-5x+1}{2x+\sqrt{5x-1}}+\dfrac{x-1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt{5x-1}}+\dfrac{1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\right)\le0\)

\(\Leftrightarrow x-1\le0\)

\(\Rightarrow\dfrac{1}{5}\le x\le1\)

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

=>\(x^2+3x-10>=0\)

=>(x+5)(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-5\end{matrix}\right.\)

\(\sqrt{x^2+5x+4}\ge2x+2\) (ĐKXĐ: \(x\ge-1\))

\(\Leftrightarrow x^2+5x+4=4x^2+8x+4\)

\(\Leftrightarrow-3x^2-3x=0\)

\(\Leftrightarrow-3x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x=0\\x+1=0\end{matrix}\right.\)

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

Vậy \(S=\left\{0;-1\right\}\)

https://meet.google.com/ais-xuwi-vhc

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow4\sqrt{2x^2-10x+16}-4x+12-4\sqrt{x-1}\le0\)

\(\Leftrightarrow4\sqrt{2x^2-10x+16}-5x+9+x+3-4\sqrt{x-1}\le0\)

\(\Leftrightarrow\frac{16\left(2x^2-10x+16\right)-\left(5x-9\right)^2}{4\sqrt{2x^2-10x+16}+5x-9}+\frac{\left(x+3\right)^2-16\left(x-1\right)}{x+3+4\sqrt{x-1}}\le0\)

\(\Leftrightarrow\frac{7\left(x-5\right)^2}{4\sqrt{2x^2-10x+16}+5x-9}+\frac{\left(x-5\right)^2}{x+3+4\sqrt{x-1}}\le0\)

\(\Leftrightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

Vậy BPT có nghiệm duy nhất \(x=5\)

ĐKXĐ: \(x\ge3\)

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

\(\Leftrightarrow x-1>2x-5+2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow4-x>2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\\left(4-x\right)^2>4\left(x^2-5x+6\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\3x^2-12x+8< 0\end{matrix}\right.\)

\(\Rightarrow\dfrac{6-2\sqrt{3}}{3}< x< \dfrac{6+2\sqrt{3}}{3}\)

Kết hợp ĐKXĐ \(\Rightarrow3\le x< \dfrac{6+2\sqrt{3}}{3}\)