ĐK: \(-\dfrac{1}{4}\le x\le3\)

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

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

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

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

Lời giải:

a. ĐKXĐ: $x\geq 0$

$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$

$\Leftrightarrow 13\sqrt{2x}=28$

$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$

$\Leftrightarrow 2x=\frac{784}{169}$

$\Leftrightarrow x=\frac{392}{169}$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$

PT $\Leftrightarrow \frac{3x-2}{x+1}=9$

$\Rightarrow 3x-2=9(x+1)$

$\Leftrightarrow x=\frac{-11}{6}$ (tm)

Ta có

\(\left\{{}\begin{matrix}\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\\\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\) \(\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\) \(\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\) \(\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4}=2\\3-\left(x-1\right)^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

a: Ta có: \(\sqrt{x^2-x+3}+7=10\)

\(\Leftrightarrow x\left(x-1\right)=0\)

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

b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)

\(\Leftrightarrow x^2-4x+8=4\)

\(\Leftrightarrow x-2=0\)

hay x=2

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

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

đến đây dễ rùi

Phương pháp giải như sau :  

Trước hết phải có ĐKXĐ là  \(x>1\)

Biến đổi phương trình về dạng \(\sqrt{\frac{5\sqrt{2}+7}{x+1}}+4\left(x+1\right)=3\left(\sqrt{2}+1\right)\)        (1)

Áp dụng bất đẳng thức AM-GM Côsi cho 3 số ta có

\(VT=\sqrt{\frac{5\sqrt{2}+7}{x+1}}+4\left(x+1\right)=\frac{\sqrt{5\sqrt{2}+7}}{2\sqrt{x+1}}+\frac{\sqrt{5\sqrt{2}+7}}{2\sqrt{x+1}+1}+4\left(x+1\right)\) \(\ge3\sqrt[3]{\frac{\sqrt{5\sqrt{2}+7}}{2\sqrt{x+1}}\cdot\frac{\sqrt{5\sqrt{2}+7}}{2\sqrt{x+1}}\cdot4\left(x+1\right)}\)\(=3\sqrt[3]{5\sqrt{2}+7}=3\sqrt[3]{\left(\sqrt{2}+1\right)^3}=3\left(\sqrt{2}+1\right)=VP\)nên

(1)   \(\Leftrightarrow\frac{\sqrt{5\sqrt{2}+7}}{2\sqrt{x+1}}=4\left(x+1\right)\Leftrightarrow x=\frac{\sqrt{2}-3}{4}\)(tm)

Kết luận:...        (Đây chỉ là hướng giải các bạn tự trình bày nhé, chúc học tốt)

\(3x-2\sqrt{4x-3}=3\) (ĐK: \(x\ge1\))

\(\Leftrightarrow2\sqrt{4x-3}=3x-3\)

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

\(\Leftrightarrow4\cdot\left(4x-3\right)=9x^2-18+9\)

\(\Leftrightarrow16x-12-9x^2+18x-9=0\)

\(\Leftrightarrow34x-9x^2-21=0\)

\(\Leftrightarrow27x+7x-9x^2-21=0\)

\(\Leftrightarrow\left(27x-9x^2\right)-\left(21-7x\right)=0\)

\(\Leftrightarrow9x\left(3-x\right)-7\left(3-x\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(9x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\9x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(n\right)\\x=\frac{7}{9}\left(l\right)\end{matrix}\right.\)

Vậy: x=3

ĐKXĐ: ...

\(\Leftrightarrow3\left(2\sqrt{x+2}+\sqrt{3-x}\right)=3x+1+4\sqrt{-x^2+x+6}\)

Đặt \(2\sqrt{x+2}+\sqrt{3-x}=t>0\)

\(\Rightarrow t^2=4\left(x+2\right)+3-x+4\sqrt{\left(x+2\right)\left(3-x\right)}=3x+11+4\sqrt{-x^2+x+6}\)

Pt trở thành:

\(3t=t^2-10\)

\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2\sqrt{x+2}+\sqrt{3-x}=5\)

Ta có: \(VT=2\sqrt{x+2}+\sqrt{3-x}\le\sqrt{\left(2^2+1^2\right)\left(x+2+3-x\right)}=5\)

\(\Rightarrow VT\le VP\)

Dấu "=" xảy ra khi và chỉ khi: \(\frac{\sqrt{x+2}}{2}=\sqrt{3-x}\Leftrightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)