ĐK: x \(\ge\)\(\frac{8}{3}\)

pt <=> \(4.\left(x-3\right)+9-3.\sqrt{5x-6}=\sqrt{3x-8}-1\)

<=> \(4.\left(x-3\right)+3.\left(3-\sqrt{5x-6}\right)=\sqrt{3x-8}-1\)

<=>  \(4.\left(x-3\right)+3.\frac{\left(3-\sqrt{5x-6}\right)\left(3+\sqrt{5x-6}\right)}{3+\sqrt{5x-6}}=\frac{\left(\sqrt{3x-8}-1\right)\left(\sqrt{3x-8}+1\right)}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+3.\frac{9-5x+6}{3+\sqrt{5x-6}}=\frac{3x-8-1}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+15.\frac{3-x}{3+\sqrt{5x-6}}-3.\frac{x-3}{\sqrt{3x-8}+1}=0\)

<=> \(\left(x-3\right)\left(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}\right)=0\)

<=> x = 3 (thoả mãn) hoặc \(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}=0\) (2)

Giải (2):  (2) <=> \(\frac{15}{6}-\frac{15}{3+\sqrt{5x-6}}+\frac{3}{2}-\frac{3}{\sqrt{3x-8}+1}=0\)

<=> \(15\left(\frac{1}{6}-\frac{1}{3+\sqrt{5x-6}}\right)+3.\left(\frac{1}{2}-\frac{1}{\sqrt{3x-8}+1}\right)=0\)

<=>  \(15.\frac{\sqrt{5x-6}-3}{6.\left(3+\sqrt{5x-6}\right)}+3.\frac{\sqrt{3x-8}-1}{2.\left(\sqrt{3x-8}+1\right)}=0\)

<=> \(15.\frac{5.\left(x-3\right)}{6.\left(3+\sqrt{5x-6}\right)^2}+3.\frac{3.\left(x-3\right)}{2.\left(\sqrt{3x-8}+1\right)^2}=0\)

<=> \(\left(x-3\right).\left(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}\right)=0\)

<=> x = 3 Vì \(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}>0\) với mọi x \(\ge\frac{8}{3}\)

Vậy pt có 1 nghiệm duy nhất x = 3

\(ĐK:x\in R\)

Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)

\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)

a) Điều kiện $x \ge -5$. Đặt $\sqrt{x+5}=a$ thì $x=a^2-5$. Thay vào ta có $$\begin{array}{l} (a^2-5)^2-7(a^2-5)=6a-30 \\ \Leftrightarrow a^4-17a^2-6a+90=0 \Leftrightarrow (a^2+6a+10)(a-3)^2=0 \end{array}$$

Vậy $a=3 \Leftrightarrow \boxed{ x= 4}$.

\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)

Đặt: \(\left\{{}\begin{matrix}a=\sqrt{3x^2+5x+8}\\b=\sqrt{3x^2+5x+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=3x^2+5x+8\\b^2=3x^2+5x+1\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}a-b=1\\a^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\\left(1+b\right)^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\1+2b+b^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x^2+5x+8}=4\\\sqrt{3x^2+5x+1}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x+8=16\\3x^2+5x+1=9\end{matrix}\right.\) \(\Leftrightarrow3x^2+5x-8=16\) \(\Leftrightarrow3x^2-3x+8x-8=0\)

\(\Leftrightarrow3x\left(x-1\right)+8\left(x-1\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(n\right)\\x=-\dfrac{8}{3}\left(n\right)\end{matrix}\right.\)

\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+6x+1}=1\)

Đặt : \(3x^2+5x+8=a\) . Phương trình trở thành :

\(\sqrt{a}-\sqrt{a-7}=1\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{a-7}\right)^2=1\)

\(\Leftrightarrow a-2\sqrt{a\left(a-7\right)}+a-7=1\)

\(\Leftrightarrow2a-2\sqrt{a\left(a-7\right)}=8\)

\(\Leftrightarrow2\sqrt{a\left(a-7\right)}=2a-8\)

\(\Leftrightarrow\sqrt{a\left(a-7\right)}=a-4\)

\(\Leftrightarrow a\left(a-7\right)=\left(a-4\right)^2\)

\(\Leftrightarrow a^2-7a=a^2-8a+16\)

\(\Leftrightarrow a=16\)

\(\Leftrightarrow3x^2+5x+8=16\)

\(\Leftrightarrow3x^2+5x-8=0\)

\(\Delta=5^2+4.3.8=25+96=121>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{121}}{6}=1\\x_2=\dfrac{-5-\sqrt{121}}{6}=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{8}{3};1\right\}\)

\(\sqrt{x-2}-3\sqrt{x^2-4}=0\left(x\ge2\right)\)

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

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

(+) x - 2 = 0

<=> x = 2 (nhận)

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

\(\Leftrightarrow9\left(x+2\right)=1\)

\(\Leftrightarrow x=\dfrac{1}{9}-2\)

\(\Leftrightarrow x=-\dfrac{17}{9}\) (loại)

a) Bình phương lên thôi

Đk: \(x\ge1\)

\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

\(\Rightarrow\left(x-1\right)+\left(5x-1\right)-2\sqrt{\left(x-1\right)\left(5x-1\right)}=3x-2\)

\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(5x-1\right)}=3x\)

\(\Leftrightarrow4\left(x-1\right)\left(5x-1\right)=9x^2\) (vì \(x\ge1\))

\(\Leftrightarrow11x^2-24x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{11}\end{matrix}\right.\)

Thử lại thấy ko thỏa mãn

Vậy pt vô nghiệm.