\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)

\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)

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

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

\(2\sqrt{x+3}=x+4\)

\(4\left(x+3\right)=x^2+8x+14\)

\(x^2+4x+2=0\)

\(\Delta=16-8=8\)

\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)

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

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

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

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

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

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

\(\Leftrightarrow x-1=0\)

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

a: ĐKXĐ: \(x\notin\left\{3;-5\right\}\)

\(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

=>\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

=>\(\dfrac{5x+25-3x+9}{15}=\dfrac{5x+25-3x+9}{\left(x-3\right)\left(x+5\right)}\)

=>(x-3)(x+5)=15

=>\(x^2+2x-15-15=0\)

=>\(x^2+2x-30=0\)

=>\(\left(x+1\right)^2=31\)

=>\(\left[{}\begin{matrix}x+1=\sqrt{31}\\x+1=-\sqrt{31}\end{matrix}\right.\Leftrightarrow x=-1\pm\sqrt{31}\left(nhận\right)\)

b: ĐKXĐ: \(x\in R\)

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

=>\(\left\{{}\begin{matrix}x^2+x+1=\left(3-x\right)^2\\x< =3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\x^2-6x+9=x^2+x+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\-7x=-8\end{matrix}\right.\Leftrightarrow x=\dfrac{8}{7}\left(nhận\right)\)

c:

ĐKXĐ: \(x\in R\)

 \(x^2-x+\sqrt{x^2-x+24}=18\)

=>\(x^2-x+24+\sqrt{x^2-x+24}=42\)

=>\(\left(\sqrt{x^2-x+24}\right)^2+\left(\sqrt{x^2-x+24}\right)-42=0\)

=>\(\left(\sqrt{x^2-x+24}+7\right)\left(\sqrt{x^2-x+24}-6\right)=0\)

=>\(\sqrt{x^2-x+24}-6=0\)

=>\(x^2-x+24=36\)

=>\(x^2-x-12=0\)

=>(x-4)(x+3)=0

=>\(\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)

b) cách khác:

\(pt\Leftrightarrow11-x-4\sqrt{x+3}-2\sqrt{3-2x}=0\)

\(\Leftrightarrow3-2x-2\sqrt{3-2x}+1+x+3-4\sqrt{x+3}+4=0\)

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

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

\(\Leftrightarrow x=1\)

b liên hợp hoặc cosi, đặt ẩn cx đc

b)\(Pt\Leftrightarrow x-1+4\sqrt{x+3}-8+2\sqrt{3-2x}-2=0\)

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

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

\(\Leftrightarrow x=1\)

Cộng 2 phương trình lại 
VT có:\(\sqrt{x}+\sqrt{32-x}\le8;\sqrt[4]{x}+\sqrt[4]{32-x}\le4\) nên VT\(\le\)12
VP có:\(y^2-6y+21=\left(y-3\right)^2+12\ge12\)
Nghiệm \(x=16;y=3\)

điều kiện: 0=<x =< 32

hệ đã cho tương đương với: \(\hept{\begin{cases}\left(\sqrt{x}+\sqrt{32-x}\right)+\left(\sqrt[4]{x}+\sqrt[4]{32-x}\right)=y^2-6y+21\\\sqrt{x}+\sqrt[4]{32-x}=y^2-3\end{cases}}\)

theo bất đẳng thức Bunhiacopsky ta có:

\(\left(\sqrt{x}+\sqrt{32-x}\right)^2\le\left(1^2+1^2\right)\left(x+32-x\right)=64\)

\(\Rightarrow\sqrt{x}+\sqrt{32-x}\le8\)

\(\left(\sqrt[4]{x}+\sqrt[4]{32-x}\right)^4\le\left[2\left(\sqrt{x}+\sqrt{32-x}\right)\right]^2\le256\Rightarrow\sqrt[4]{x}+\sqrt[4]{32-x}\le4\)

\(\Rightarrow\left(\sqrt{x}+\sqrt{32-x}\right)+\left(\sqrt[4]{x}+\sqrt[4]{32-x}\right)\le12\)

mặt khác \(y^2-6y+21=\left(y-3\right)^2+12\ge12\)

đẳng thức xảy ra khi x=16 và y=3 (tm)

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

ĐKXĐ: \(2059-x\ge0\)

PT đã cho tương đương với:

\(\sqrt{2059-x}+\sqrt{2059-x+2994}+\sqrt{2059-x+95}=24\)(*)

Mà VT của pt(*)\(\ge0+\sqrt{2994}+\sqrt{95}>24=VP\) nên pt(*) vô nghiệm

Vậy pt đã cho vô nghiệm