ĐKXĐ: \(-\frac{16}{3}\le x\le4\)

\(\Leftrightarrow3x^2-12x+36=12\sqrt{4-x}+3\sqrt{3x+16}\)

\(\Leftrightarrow3x^2-9x+4\left(6-x-3\sqrt{4-x}\right)+\left(x+12-3\sqrt{3x+16}\right)=0\)

\(\Leftrightarrow3\left(x^2-3x\right)+\frac{4\left(x^2-3x\right)}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+12+3\sqrt{3x+16}}=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(3+\frac{4}{6-x+3\sqrt{4-x}}+\frac{1}{x+12+3\sqrt{3x+16}}\right)=0\)

\(\Leftrightarrow x^2-3x=0\)

ĐLXĐ:\(x\ge-1\)

\(\sqrt{x^2+4x+12}=2x-4+\sqrt{x+1}\)

\(\Leftrightarrow\left[\sqrt{x^2+4x+12}-\left(6-3x\right)\right]-\left[\sqrt{x+1}-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\frac{x^2+4x+12-36+36x-9x^2}{\sqrt{x^2+4x+12}+2-3x}-\frac{x+1-x^2+4x-4}{\sqrt{x+1}+x+2}=0\)

\(\Leftrightarrow\frac{-8x^2+40x-24}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\frac{8\left(-x^2+5x-3\right)}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\left(-x^2+5x-3\right)\left[\frac{8}{\sqrt{x^2+4x+12}+2-3x}-\frac{1}{\sqrt{x+1}+x-2}\right]=0\)

TH1:\(-x^2+5x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

TH2:........ ( chắc vô nghiệm )

phần mẫu phải là \(\sqrt{x^2+4x+12}+6-3x\) chứ :vv Hơi lỗi nhưng cảm ơn nhé !!

ko biết

Bài quá dễ tự làm đi 

k mình mình giải cho

ĐKXĐ: -16/3<=x<=4

\(\)\(x^2-4x+12=4\sqrt{4-x}+\sqrt{3x+16}\)

=>\(x^2-4x+12-\left(-x+12\right)=4\sqrt{4-x}-\left(-\frac43x+8\right)+\sqrt{3x+16}-\left(\frac13x+4\right)\)

=>\(x^2-3x=\frac43\left\lbrack3\sqrt{4-x}-\left(6-x\right)\right\rbrack+\frac13\left\lbrack3\cdot\sqrt{3x+16}-\left(x+12\right)\right\rbrack\)

=>\(x^2-3x=\frac43\cdot\frac{9\left(4-x\right)-\left(6-x\right)^2}{3\sqrt{4-x}+\left(6-x\right)}+\frac13\cdot\frac{9\left(3x+16\right)-\left(x+12\right)^2}{3\sqrt{3x+16}+\left(x+12\right)}\)

=>\(x^2-3x=\frac43\cdot\frac{36-9x-x^2+12x-36}{3\sqrt{4-x}+\left(6-x\right)}+\frac13\cdot\frac{27x+144-x^2-24x-144}{3\sqrt{3x+16}+\left(x+12\right)}\)

=>\(\left(x^2-3x\right)\left(1+\frac43\cdot\frac{1}{3\sqrt{4-x}+6-x}+\frac{1}{3\left(3\sqrt{3x+16}+x+12\right)}\right)=0\)

=>\(x^2-3x=0\)

=>x(x-3)=0

=>x=0(nhận) hoặc x=3(nhận)

đồng đội toán còn đi hỏi đây

ban giai ho to

b,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}-16\sqrt{x+1}=0\) (dk \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}\left(4-3+2-16\right)=0\)

\(\Leftrightarrow\sqrt{x+1}.-13=0\)

\(\Leftrightarrow x=-1\)

<=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2=5\)

mà \(\sqrt{3\left(x+1\right)^2+9}\ge3\), \(\sqrt{5\left(x^2-1\right)^2+4}\ge4\), \(2\left(x+1\right)^2\ge0\)với mọi x 

=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2\ge3+2+0=5\)

'=" xảy ra<=> x+1=0<=> x=-1