\(ĐKXĐ:0\le x\le6\)

\(\Leftrightarrow\sqrt{6x-x^2}-2\left(6x-x^2\right)+15=0\)

Đặt \(\sqrt{6x-x^2}=t\left(t\ge0\right)\)

PT trở thành:

\(2t^2-t-15=0\)

\(\Leftrightarrow\left(t-3\right)\left(2t+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=\frac{-5}{2}\end{cases}}\)

\(TH1:t=3\Rightarrow\sqrt{6x-x^2}=3\Rightarrow6x-x^2=9\)

\(\Leftrightarrow x^2-6x+9=0\)

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

\(\Leftrightarrow x=3\)

\(TH2:t=\frac{-5}{2}\)không TMĐK \(t\ge0\)

Vậy PT có nghiệm là \(S=\left\{3\right\}\)

a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

đôi lúc là +- 10 thì sao 

ta có : \(\left(m^2+1\right)x^2-\left(2-m\right)=0\Rightarrow2-m=\left(m^2+1\right)x^2\ge1\)

VẬY PT CÓ NGHIỆM KHI  \(2-m\ge1\Leftrightarrow m\le1\).

\(\Rightarrow x^2=\frac{2-m}{m^2+1}\Leftrightarrow x=\sqrt{\frac{2-m}{m^2+1}}\)hoặc x=\(-\sqrt{\frac{2-m}{m^2+1}}\)

\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)

Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)

Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)

\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0

⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x²+12x+9=0

⇔ x⁴+ 4x³+2x²-6x²+12x+9=0

⇔ x²+4x³-4x² +12x=-9

⇔x²+ 4x(x-x+3)=-9

⇔x²+12x=-9

⇔x(x+12)=-9

⇔ {x=-9 hoặc x+12=-9}

⇔ {x=-9 hoặc x=-21}

S={-9;-21}