a)\(=>\left[{}\begin{matrix}3x-2=0\\2x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}3x=2\\2x=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

b.\(\dfrac{x-1}{x+1}+\dfrac{4}{1-x^2}=\dfrac{2\left(x+1\right)}{x-1}\)

\(ĐK:x\ne\pm1\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-1\right)-4}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

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

\(\Leftrightarrow x^2-2x+1-4=2x^2+4x+2\)

\(\Leftrightarrow x^2+6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

=>[3x−2=02x+5=0=>[3x=22x=−5=>⎡⎢ ⎢⎣x=23x=−52

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

b. 2 + \(\sqrt{2x-1}=x\)       ĐKXĐ: \(x\ge0,5\)

<=> \(\sqrt{2x-1}\) = x - 2

<=> 2x - 1 = (x - 2)²

<=> 2x - 1 = x² - 4x + 4

<=> -x² + 2x + 4x - 4 - 1 = 0

<=> -x² + 6x - 5 = 0

<=> -x² + 5x + x - 5 = 0

<=> -(-x² + 5x + x - 5) = 0

<=> x² - 5x - x + 5 = 0

<=> x(x - 5) - (x - 5) = 0

<=> (x - 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

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)

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 \(x=2\)