\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\\ \Leftrightarrow\left(12x+7\right)^2\left(12x+8\right)\left(12x+6\right)=72\)

\(\text{Đặt}:12x+7=t\)

\(\Leftrightarrow t^2\left(t+1\right)\left(t-1\right)=72\\ \Leftrightarrow t^2\left(t^2-1\right)=72\\ \Leftrightarrow t^2\left(t^2-1\right)-72=0\\ \Leftrightarrow t^4-t^2-72=0\\ \Leftrightarrow t^4-9t^2+8t^2-72=0\\ \Leftrightarrow t^2\left(t^2-9\right)+8\left(t^2-9\right)=0\\ \Leftrightarrow\left(t^2-9\right)\left(t^2-8\right)=0\\ \Leftrightarrow\left(t-3\right)\left(t+3\right)=0\left(do:t^2-8\ne0\right)\\ \Leftrightarrow\left(12x+7+3\right)\left(12x+7-3\right)=0\\ \Leftrightarrow\left(12x+10\right)\left(12x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}12x+10=0\\12x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}12x=-10\\12x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{6}\\x=-\frac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)-4=0\)

\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4=0\)

Đặt \(12x^2+11x-1=a\)

\(\left(a+3\right)a-4=0\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}12x^2+11x-1=1\\12x^2+11x-1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}12x^2+11x-2=0\\12x^2+11x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)

ĐK: \(-\dfrac{1}{4}\le x\le3\)

\(pt\Leftrightarrow4x+1-6\sqrt{4x+1}+9+3-x-2\sqrt{3-x}+1=0\)

\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

<=> (12²x²+2.12.7x + 7²).(6x²+7x+2) = 3 
<=> [24.(6x² +7x +2) +1].(6x² +7x +2) =3 
đặt: a= 6x² +7x +2 
<=> (24a+1).a = 3 
=> a=...

⇔ (3a-1)(8a+3)=0 
⇔ a=1/3 hoặc a=−3/8 

\(x=0\) không phải nghiệm, pt tương đương:

\(\frac{12}{x+4+\frac{2}{x}}-\frac{3}{x+2+\frac{2}{x}}=1\)

Đặt \(x+2+\frac{2}{x}=a\)

\(\frac{12}{a+2}-\frac{3}{a}=1\Leftrightarrow12a-3\left(a+2\right)=a\left(a+2\right)\)

\(\Leftrightarrow a^2-7a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2+\frac{2}{x}=1\\x+2+\frac{2}{x}=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-4x+2=0\end{matrix}\right.\)

\(ĐK:\frac{3-\sqrt{17}}{2}\le x\le\frac{3+\sqrt{17}}{2};\orbr{\begin{cases}x\ge\frac{1}{\sqrt{5}}\\x\le-\frac{1}{\sqrt{5}}\end{cases}}\)

Bình phương hai vế của phương trình, ta được: \(2-x^2+3x=5x^2-1\Leftrightarrow6x^2-3x-3=0\Leftrightarrow3\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\left(tmđk\right)\)

Vậy phương trình có tập nghiệm S = {1; ^-1/_2} }

Lời giải:

ĐK: $x\geq \frac{-1}{3}$. Ta có:

\(4x^2+5+\sqrt{3x+1}=13x\)

\(\Leftrightarrow (4x^2-11x+3)-(2x-2-\sqrt{3x+1})=0(*)\)

TH1: Nếu \(2x-2+\sqrt{3x+1}=0(1)\)

\(\Rightarrow \sqrt{3x+1}=2-2x\Rightarrow \left\{\begin{matrix} x\leq 1\\ 3x+1=(2-2x)^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 1\\ 4x^2-11x+3=0\end{matrix}\right.\Rightarrow x=\frac{11-\sqrt{73}}{8}\) . Thử lại vào PT ban đầu không thấy đúng (loại)

TH2: Nếu $2x-2+\sqrt{3x+1}\neq 0$ (tức là \(x\neq \frac{11-\sqrt{73}}{8}\))

\((*)\Leftrightarrow (4x^2-11x+3)-\frac{(2x-2)^2-(3x+1)}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow (4x^2-11x+3)-\frac{4x^2-11x+3}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow \frac{(4x^2-11x+3)(2x-3+\sqrt{3x+1})}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow \left[\begin{matrix} 4x^2-11x+3=0\\ 2x-3+\sqrt{3x+1}=0\end{matrix}\right.\)

Nếu $4x^2-11x+3=0\Rightarrow x=\frac{11+\sqrt{73}}{8}$ (loại TH $x=\frac{11-\sqrt{73}}{8}$

Nếu \(2x-3+\sqrt{3x+1}=0\Rightarrow \left\{\begin{matrix} x\leq \frac{3}{2}\\ (2x-3)^2=3x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{3}{2}\\ 4x^2-15x+8=0\end{matrix}\right.\Rightarrow x=\frac{15-\sqrt{97}}{8}\)

Thử lại thấy thỏa mãn. Vậy.........

Lời giải:

ĐK: $x\geq \frac{-1}{3}$. Ta có:

\(4x^2+5+\sqrt{3x+1}=13x\)

\(\Leftrightarrow (4x^2-11x+3)-(2x-2-\sqrt{3x+1})=0(*)\)

TH1: Nếu \(2x-2+\sqrt{3x+1}=0(1)\)

\(\Rightarrow \sqrt{3x+1}=2-2x\Rightarrow \left\{\begin{matrix} x\leq 1\\ 3x+1=(2-2x)^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 1\\ 4x^2-11x+3=0\end{matrix}\right.\Rightarrow x=\frac{11-\sqrt{73}}{8}\) . Thử lại vào PT ban đầu không thấy đúng (loại)

TH2: Nếu $2x-2+\sqrt{3x+1}\neq 0$ (tức là \(x\neq \frac{11-\sqrt{73}}{8}\))

\((*)\Leftrightarrow (4x^2-11x+3)-\frac{(2x-2)^2-(3x+1)}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow (4x^2-11x+3)-\frac{4x^2-11x+3}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow \frac{(4x^2-11x+3)(2x-3+\sqrt{3x+1})}{2x-2+\sqrt{3x+1}}=0\)

\(\Leftrightarrow \left[\begin{matrix} 4x^2-11x+3=0\\ 2x-3+\sqrt{3x+1}=0\end{matrix}\right.\)

Nếu $4x^2-11x+3=0\Rightarrow x=\frac{11+\sqrt{73}}{8}$ (loại TH $x=\frac{11-\sqrt{73}}{8}$

Nếu \(2x-3+\sqrt{3x+1}=0\Rightarrow \left\{\begin{matrix} x\leq \frac{3}{2}\\ (2x-3)^2=3x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{3}{2}\\ 4x^2-15x+8=0\end{matrix}\right.\Rightarrow x=\frac{15-\sqrt{97}}{8}\)

Thử lại thấy thỏa mãn. Vậy.........