ĐKXĐ: \(x>\dfrac{1}{5}\)

\(1-3x^2< \left(x+2\right)\sqrt[]{5x-1}+5x-1\)

\(\Leftrightarrow3x^2+5x-2+\left(x+2\right)\sqrt{5x-1}\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-1\right)+\left(x+2\right)\sqrt{5x-1}>0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-1+\sqrt{5x-1}\right)>0\)

\(\Leftrightarrow3x-1+\sqrt{5x-1}>0\)

\(\Leftrightarrow\sqrt{5x-1}>1-3x\)

TH1: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{5}\\1-3x< 0\end{matrix}\right.\) \(\Leftrightarrow x>\dfrac{1}{3}\)

TH2: \(\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\5x-1>9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\9x^2-11x+2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{9}< x\le\dfrac{1}{3}\)

Kết luận: \(x>\dfrac{2}{9}\)

Đặt \(t=3x^2+5x+2\)

Do đó ta có:\(\sqrt{3x^2+5x+7}-\sqrt{3x^2+5^2+2}=1\)

               \(\sqrt{t+5}-\sqrt{t}=1\)

                 \(\left(\sqrt{t+5}-\sqrt{t}\right)^2=1\)

                 \(t+5-2\sqrt{t\left(t+5\right)}+t=1\)

                \(2t-2\sqrt{t\left(t+5\right)}+5=1\)

                \(2t+4=2\sqrt{t\left(t+5\right)}\)

                 \(\left(t+2\right)^2=t\left(t+5\right)\)

                      \(4t+4=5t\)

                            \(\Rightarrow t=4\)

Tại t=4 ta được:\(3x^2+5x+2=4\)

                        \(3x^2+5x-2=0\)

                        \(3x^2+6x-x-2=0\)

                               \(\Rightarrow\left(3x-1\right)\left(x+2\right)=0\)

               \(\Rightarrow\orbr{\begin{cases}3x-1=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

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

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

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

\(\Leftrightarrow x-1\le0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x-7=3x+14\\x\ge-\dfrac{14}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x^2+2x-21=0\\x\ge-\dfrac{14}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+3\right)\left(3x-7\right)=0\\x\ge-\dfrac{14}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{7}{3}\end{matrix}\right.\)

a/ ĐKXĐ: \(x\ge-9\)

- Với \(-9\le x< 9\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng

- Với \(x\ge9\Rightarrow\sqrt{x+9}>4\Rightarrow4-\sqrt{x+9}< 0\)

BPT tương đương:

\(7\left(\sqrt{x+9}-4\right)>x-9\)

Đặt \(\sqrt{x+9}=t\ge3\sqrt{2}\)

\(\Rightarrow7t-28>t^2-18\)

\(\Leftrightarrow t^2-7t+10< 0\Leftrightarrow2< t< 5\)

\(\Rightarrow\sqrt{x+9}< 5\) \(\Leftrightarrow x< 16\)

Kết hợp lại ta được nghiệm của BPT là:

\(-9\le x< 16\)

b/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-\frac{2}{3}\\x\le-1\end{matrix}\right.\)

Đặt \(3x^2+5x+2=t\ge0\)

\(\Leftrightarrow\sqrt{t+5}-\sqrt{t}>1\)

\(\Leftrightarrow\sqrt{t+5}>\sqrt{t}+1\)

\(\Leftrightarrow t+5>t+1+2\sqrt{t}\)

\(\Leftrightarrow\sqrt{t}< 2\Rightarrow t< 4\)

\(\Rightarrow3x^2+5x+2< 4\)

\(\Leftrightarrow3x^2+5x-2< 0\) \(\Rightarrow-2< x< \frac{1}{3}\)

Kết hợp ĐKXĐ ta được nghiệm của BPT:

\(\left[{}\begin{matrix}-2< x\le-1\\-\frac{2}{3}\le x< \frac{1}{3}\end{matrix}\right.\)

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)