Dat \(\sqrt{x+8}=a,\sqrt{x+3}=b\)

=> a.b=\(\sqrt{x^2+11x+24},a^2-b^2=5\)

pt<=> (a-b)(ab+1)=a²-b²

=> (a-b)(ab+1)=(a-b)(a+b)

=> (a-b)(ab+1)-(a-b)(a+b)=0

=> (a-b)(ab+1-a-b)=0

=> (a-b)[a(b-1)-(b-1)]=0

=> (a-b)(a-1)(b-1)=0

=> \(\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)

Voi a=b thi : x+8=x+3

=> pt vo nghiem

Voi a=1 thi x+8=1 => x=-7

Voi b=1 thi x+3=1 => x=-2

Điều kiện: x\(\ge\) -3

PT <=>  \(\left(\sqrt{x+8}+\sqrt{x+3}\right)\left(\sqrt{x+8}-\sqrt{x+3}\right)\left(\sqrt{x^2+11x+24}+1\right)=5\left(\sqrt{x+8}+\sqrt{x+3}\right)\)

<=> \(\left(x+8-x-3\right)\left(\sqrt{x^2+11x+24}+1\right)=5\left(\sqrt{x+8}+\sqrt{x+3}\right)\)

<=> \(\sqrt{\left(x+3\right)\left(x+8\right)}+1=\sqrt{x+8}+\sqrt{x+3}\)

<=>   \(\left(\sqrt{\left(x+3\right)\left(x+8\right)}-\sqrt{x+8}\right)+\left(1-\sqrt{x+3}\right)=0\)

<=> \(\left(1-\sqrt{x+8}\right).\left(1-\sqrt{x+3}\right)=0\)

<=>  \(\sqrt{x+8}=1\) hoặc \(\sqrt{x+3}=1\)

<=> x+ 8 = 1 hoặc x + 3 = 1

<=> x = -7 hoặc x = - 2

Đối chiếu Đk => x = - 2 là nghiệm của PT

ĐK: \(x\ge-\dfrac{5}{2}\)

\(\Leftrightarrow3x^2-4x-4=2x+5\)

\(\Leftrightarrow3x^2-6x-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) (thỏa mãn)

b.

ĐKXĐ: \(3\le x\le8\)

\(\Leftrightarrow-x^2+11x-24-\sqrt{-x^2+11x-24}-2=0\)

Đặt \(\sqrt{-x^2+11x-24}=t\ge0\)

\(\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{-x^2+11x-24}=2\)

\(\Leftrightarrow-x^2+11x-28=0\Rightarrow\left[{}\begin{matrix}x=7\\x=4\end{matrix}\right.\)

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)