\(\sqrt{4+20x}=3x+2\left(x\ge-\dfrac{1}{5}\right)\\ \Leftrightarrow4+20x=9x^2+12x+4\\ \Leftrightarrow9x^2-8x=0\\ \Leftrightarrow x\left(9x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=\dfrac{8}{9}\left(N\right)\end{matrix}\right.\\ \sqrt{2x+5}=x+1\left(x\ge-\dfrac{5}{2}\right)\\ \Leftrightarrow2x+5=x^2+2x+1\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-2\left(N\right)\end{matrix}\right.\)

\(\sqrt{4+20x}=3x+2\\ \Leftrightarrow4+20x=\left(3x+2\right)^2\\ \Leftrightarrow4+20x=9x^2+12x+4\\ \Leftrightarrow-4-20x+9x^2+12x+4=0\\ \Leftrightarrow9x^2-8x=0\\ \Leftrightarrow x\left(9x-8\right)=0\\ \Leftrightarrow x=0hoặcx=\dfrac{8}{9}\)

\(\sqrt{2x+5}=x+1\\ \Leftrightarrow2x+5=\left(x+1\right)^2\\ \Leftrightarrow2x+5=x^2+2x+1\\ \Leftrightarrow x^2+2x+1-2x-5=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)

a: Ta có: \(\sqrt{20x+4}=3x+2\)

\(\Leftrightarrow9x^2+12x+4=20x+4\)

\(\Leftrightarrow9x^2-8x=0\)

\(\Leftrightarrow x\left(9x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{8}{9}\left(nhận\right)\end{matrix}\right.\)

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

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

\(\sqrt{3x^2-12x+21}=\sqrt{3x^2-12x+12+9}=\sqrt{3\left(x-2\right)^2+9}\ge\sqrt{9}=3\)

\(\sqrt{5x^2-20x+24}=\sqrt{5x^2-20x+20+4}=\sqrt{5\left(x-2\right)^2+4}\ge\sqrt{4}=2\)

\(-2x^2+8x-3=-2x+8x-8+5=-2\left(x-2\right)^2+5\le5\)

\(VP\ge3+2=5,VT\le5\)

Suy ra \(VP=VT=5\)

Suy ra nghiệm của phương trình đạt tại \(x-2=0\Leftrightarrow x=2\).

câu trả lời là : ko bt =))

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: \(x\ge\dfrac{1}{5}\)

\(\Leftrightarrow\sqrt{3x+5}-\sqrt{2x+6}+\sqrt{5x-1}-2=0\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{3x+5}+\sqrt{2x+6}}+\dfrac{5\left(x-1\right)}{\sqrt{5x-1}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{3x+5}+\sqrt{2x+6}}+\dfrac{5}{\sqrt{5x-1}+2}\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)
 

ĐKXĐ: \(\frac{4-\sqrt{10}}{2}\le x\le\frac{4+\sqrt{10}}{2}\)

Đặt : \(\sqrt{3x^2-12x+21}=a;\sqrt{5x^2-20x+24}=b\left(a,b>0\right)\Rightarrow a^2-b^2=-2x^2+8x-3\)

Khi đó pt trở  thành:

\(a+b=a^2-b^2\)

\(\Rightarrow a=b\)

Theo cách đặt:   \(\sqrt{3x^2-12x+21}=\sqrt{5x^2-20x+24}\)

\(\Leftrightarrow2x^2-8x+3=0\)

Đến đây bạn tự giải nha