ĐKXĐ : \(\left\{{}\begin{matrix}2x^2+5x-3\ge0\\2x-1\ge0\\x+3\ge0\end{matrix}\right.\)

Phương trình tương đương :  \(2x-2\sqrt{2x^2+5x+3}=1+x.\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\) 

\(\Leftrightarrow2x-1-2\sqrt{\left(2x-1\right)\left(x+3\right)}=x.\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=x\\\sqrt{2x-1}=2\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{2}\left(\text{loại}\right)\end{matrix}\right.\)

Vậy x = 1 là nghiệm phương trình

b )  ĐKXĐ : x ≥  2 

\(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\) 

\(\sqrt{4.\left(x-2\right)}-\dfrac{12}{3}\sqrt{x-2}=-1\)

\(2\sqrt{x-2}-4\sqrt{x-2}=-1\)

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

      \(\sqrt{x-2}=\dfrac{1}{2}\)

       \(\left(\sqrt{x-2}\right)^2=\left(\dfrac{1}{2}\right)^2\)

        \(x-2=\dfrac{1}{4}\) 

        \(x=\dfrac{1}{8}\)

Vậy biểu thức vô nghiệm

a ) \(\sqrt{4-3x}=8\) 

      \(\left(\sqrt{4-3x}\right)^2=8^2\)

        \(4-3x=8\)

        \(-3x=4\)

            \(x=-\dfrac{4}{3}\) 

a ) \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)

= \(2\sqrt{9.5}+\sqrt{5}-3\sqrt{16.5}\) \

= \(2.3\sqrt{5}+\sqrt{5}-3.4\sqrt{5}\)

= \(6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)

= \(\left(6+1-12\right)\sqrt{5}\)

= \(-5\sqrt{5}\) 

b ) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)

= / \(2-\sqrt{3}\) / \(+\dfrac{2.\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}-6\sqrt{\dfrac{48}{3^2}}\)

= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{\sqrt{3}^2-1^2}-\dfrac{6}{3}\sqrt{48}\) 

= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{3-1}-2\sqrt{48}\)

=\(2-\sqrt{3}+\sqrt{3}-1-2\sqrt{16.3}\) 

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

=  \(1-8\sqrt{3}\)

ý c ) em không biết làm ☹

Đề sai. Cho $a=2; b=1$ đều không chia hết cho $5$ 

$a^5-b^5=2^5-1=31$ không chia hết cho $5$

Lời giải:
ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow (x-2\sqrt{x}+1)+\sqrt{x-1}=0$

$\Leftrightarrow (\sqrt{x}-1)^2+\sqrt{x-1}=0$

Ta thấy: $(\sqrt{x}-1)^2\geq 0; \sqrt{x-1}\geq 0$ với mọi $x\geq 1$

Do đó để tồng trên bằng $0$ thì:
$(\sqrt{x}-1)^2=\sqrt{x-1}=0$

$\Leftrightarrow x=1$ (thỏa mãn).

\(P=\dfrac{2}{\sqrt{6+4\sqrt{2}}+2}\\ =\dfrac{2}{\sqrt{\left(2\right)^2+2.2\sqrt{2}+\left(\sqrt{2}\right)^2}+2}\\= \dfrac{2}{\sqrt{\left(2+\sqrt{2}\right)^2}+2}\\ =\dfrac{2}{2+\sqrt{2}+2}\\ =\dfrac{2}{4+\sqrt{2}}\)