\(PT\Leftrightarrow\sqrt{3x+1}=\sqrt{x+4}+1\\ \Leftrightarrow3x+1=x+5+2\sqrt{x+4}\\ \Leftrightarrow2x-4=2\sqrt{x+4}\\ \Leftrightarrow x-2=\sqrt{x+4}\\ \Leftrightarrow x^2-4x+4=x+4\\ \Leftrightarrow x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Thử lại ta thấy x=0 ko thỏa mãn

Vậy PT có nghiệm x=5

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\sqrt{3x+1}=1+\sqrt{x+4}\)

\(\Leftrightarrow3x+1=1+x+4+2\sqrt{x+2}\)

\(\Leftrightarrow x+2-\sqrt{x+2}-4=0\)

Đặt \(\sqrt{x+2}=t\ge0\)

\(\Rightarrow t^2-t-4=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\\t=\dfrac{1-\sqrt{17}}{2}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+2}=\dfrac{1+\sqrt{17}}{2}\)

\(\Rightarrow x=\dfrac{5+\sqrt{17}}{2}\)

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{4}$

PT $\Leftrightarrow \sqrt{3x+4}=\sqrt{12x+7}+\sqrt{4x+1}$

$\Rightarrow 3x+4=12x+7+4x+1+2\sqrt{(12x+7)(4x+1)}$

$\Leftrightarrow 13x+4+2\sqrt{(12x+7)(4x+1)}=0$

Với $x\geq \frac{-1}{4}$ dễ thấy vế trái $>0$. Do đó pt đã cho vô nghiệm.

ĐK: \(-1\le x\le4\)

\(\sqrt{x+1}+\sqrt{4-x}=t\left(\sqrt{5}\le t\le\sqrt{10}\right)\Rightarrow\sqrt{-x^2+3x+4}=\dfrac{t^2-5}{2}\)

\(pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\)

\(\Leftrightarrow t^2+2t-15=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+5\right)=0\)

\(\Leftrightarrow t=3\left(\text{Vì }\sqrt{5}\le t\le\sqrt{10}\right)\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}=3\)

\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)

\(\Leftrightarrow\sqrt{-x^2+3x+4}=2\)

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

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

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{3x+1}-\sqrt{2x+3}=\dfrac{x-2}{4}\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{x-2}{4}\)

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

Xét (1) \(\Leftrightarrow\sqrt{3x+1}+\sqrt{2x+3}=4\)

\(\Leftrightarrow5x+4+2\sqrt{6x^2+11x+3}=16\)

\(\Leftrightarrow2\sqrt{6x^2+11x+3}=12-5x\) (\(x\le\dfrac{12}{5}\))

\(\Leftrightarrow4\left(6x^2+11x+3\right)=\left(12-5x\right)^2\)

\(\Leftrightarrow x^2-164x+132=0\Rightarrow\left[{}\begin{matrix}x=82-8\sqrt{103}\\x=82+8\sqrt{103}>\dfrac{12}{5}\left(loại\right)\end{matrix}\right.\)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le4\)

\(\Leftrightarrow x+5=\sqrt{3x+1}+2\sqrt{4-x}\)

Ta có: 

\(VP=1.\sqrt{3x+1}+2.\sqrt{4-x}\le\dfrac{1}{2}\left(1+3x+1\right)+\dfrac{1}{2}\left(4+4-x\right)=x+5\)

\(\Rightarrow VP\le VT\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\sqrt{3x+1}=1\\\sqrt{4-x}=2\end{matrix}\right.\) \(\Leftrightarrow x=0\)

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