Đặt \(\sqrt{2x^3+7}=a\)

=>6ax=3a^2+1+2x-4a

=>a=2x+1 hoặc a=1/3

=>2x^3+7=(2x+1)^2 hoặc 2x^3+7=1/3

=>\(x\in\left\{1;\dfrac{1-\sqrt{13}}{2};\sqrt[3]{-\dfrac{31}{9}}\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x+7>=0\\-2x+3>=0\end{matrix}\right.\Leftrightarrow-\dfrac{7}{2}< =x< =\dfrac{3}{2}\)

PT\(\Leftrightarrow x^2-1+2\sqrt{2x+7}=2\sqrt{-2x+3}+4\)

=>\(\left(x-1\right)\left(x+1\right)+2\sqrt{2x+7}-6=2\sqrt{-2x+3}-2\)

=>\(\left(x-1\right)\left(x+1\right)+2\cdot\dfrac{2x+7-9}{\sqrt{2x+7}+3}=2\cdot\dfrac{-2x+3-1}{\sqrt{-2x+3}+1}\)

=>\(\left(x-1\right)\left(x+1\right)+\dfrac{4\left(x-1\right)}{\sqrt{2x+7}+3}-2\cdot\dfrac{-2\left(x-1\right)}{\sqrt{-2x+3}+1}=0\)

=>\(\left(x-1\right)\left(x+1+\dfrac{4}{\sqrt{2x+7}+3}+\dfrac{4}{\sqrt{-2x+3}+1}\right)=0\)

=>x-1=0

=>x=1(nhận)

ĐK:\(x\ge\dfrac{5}{2}\)

Ta có:\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)

    \(\Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=7.2\)

    \(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+6}=14\)

    \(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

    \(\Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)

    \(\Leftrightarrow2\sqrt{2x-5}=10\)

    \(\Leftrightarrow\sqrt{2x-5}=5\)

    \(\Leftrightarrow2x-5=25\Leftrightarrow2x=30\Leftrightarrow x=15\left(tm\right)\)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)

\(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+3}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

\(\Leftrightarrow2.\sqrt{2x-5}+4=14\)

\(\Leftrightarrow\sqrt{2x-5}=5\)

\(\Leftrightarrow x=15\)

ĐK

\(\left\{{}\begin{matrix}x+3\ge0\\7-x\ge0\\2x-8\ge0\end{matrix}\right.\)

Giải hệ bất PT trên được ĐK tổng hợp là \(4\le x\le7\)

Bình phương 2 vế PT

\(x+3+7-x-2\sqrt{\left(x+3\right)\left(7-x\right)}=2x-8\)

\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(7-x\right)}=18-2x\)

BP 3 vế PT

\(4\left(x+3\right)\left(7-x\right)=324+4x^2-72x\)

\(\Leftrightarrow28x-4x^2+84-12x=324+4x^2-72x\)

\(\Leftrightarrow8x^2-88x+240=0\Leftrightarrow x^2-11x+30=0\)

Giải PT bậc 2 rồi đối chiếu với đk, bạn tự làm nốt nhé

Đặt \(\sqrt{2x-1}=a;\sqrt{3-y}=b\)

Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{5}{b}=7\\\dfrac{3}{a}-\dfrac{7}{b}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{a}+\dfrac{15}{b}=21\\\dfrac{3}{a}-\dfrac{7}{b}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{22}{b}=22\\\dfrac{1}{a}+\dfrac{5}{b}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3-y=1\\2x-1=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=\dfrac{5}{8}\end{matrix}\right.\)

Lời giải:
ĐKXĐ: $x\geq -3,5$

PT \(\Leftrightarrow (\sqrt{2x+7}-1)+(\sqrt[3]{x+4}-1)+(x^2+8x+15)=0\)

\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x+7}+1}+\frac{x+3}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+3)(x+5)=0\)

\(\Leftrightarrow (x+3)\left[\frac{2}{\sqrt{2x+7}+1}+\frac{1}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+5)\right]=0\)

Với $x\geq -3,5$ dễ thấy biểu thức trong ngoặc vuông $>0$

Do đó: $x+3=0$

$\Leftrightarrow x=-3$ (thỏa mãn)