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

a,Để \(\sqrt{x^2-8x-9}\) có nghĩ thì

 \(x^2-8x-9\ge0\)

\(\Leftrightarrow x^2+x-9x-9\ge0\)

\(\Leftrightarrow x\left(x+1\right)-9\left(x+1\right)\ge0\)

\(\Leftrightarrow\left(x+1\right)\left(x-9\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-1\\x\ge9\end{cases}\Rightarrow}x\ge9\)

\(or\orbr{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\le-1\\x\le9\end{cases}\Rightarrow}x\le-1\)

\(Để\sqrt{4-9x^2}\text{có nghĩa}\)

\(\Rightarrow4-9x^2\ge0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)\ge0\)

\(\Leftrightarrow-\frac{2}{3}\le x\le\frac{2}{3}\)

c,Để \(\sqrt{\frac{2x-3}{2x^2+1}}\)có nghĩa thì

 \(\Rightarrow\orbr{\begin{cases}2x-3\ge0\\2x^2+1>0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x\ge3\\2x^2>-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\\x^2>-\frac{1}{2}\text{(luôn đúng)}\end{cases}}\)

\(\Rightarrow x\ge\frac{3}{2}\)

\(a,\sqrt{x^2-8x+18}=\sqrt{x^2-8x+16+2}\)

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

Vì \(\left(x-4\right)^2+2>0\)với \(\forall x\)

\(\Rightarrow\)Biểu thức luôn được xác định với mọi x 

\(b,\sqrt{\frac{3x+4}{x-2}}\)

\(btxđ\Leftrightarrow\hept{\begin{cases}x-2\ne0\\\frac{3x+4}{x-2}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\\frac{3x+4}{x-2}\ge0\end{cases}}}\)

\(\frac{3x+4}{x-2}\ge0\)\(\Rightarrow\orbr{\begin{cases}3x+4\ge0;x-2\ge0\\3x+4< 0;x-2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge-\frac{4}{3};x\ge2\\x< -\frac{4}{3};x< 2\end{cases}\Rightarrow\orbr{\begin{cases}x\ge2\\x< -\frac{4}{3}\end{cases}}}\)

Mà \(x\ne2\)\(\Rightarrow x>2\)hoặc \(x< -\frac{4}{3}\)

a,\(\sqrt{x^2-8x+18=\sqrt{x^2}-8x+16+2.}\)

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

Vì \(\left(x-4\right)^2+2>0\)với\(\forall x\)

\(\Rightarrow\)Biểu thức luônđược xác định với mọi x

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

\(btxđ\Leftrightarrow\hept{\begin{cases}x-2\ne0\\\frac{3x+4}{x-2}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\\frac{3x+4}{x-2}\ge0\end{cases}}}\)

\(\frac{3x+4}{x-2}\ge0\Rightarrow\orbr{\begin{cases}3x+4\ge0;x-2\ge0\\3x+4< 0;x-2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge-\frac{4}{3};x\ge2\\x< \ge-\frac{4}{3};x< 2\end{cases}\Rightarrow\orbr{\begin{cases}x\ge2\\x< -\frac{4}{3}\end{cases}}}\)

\(\Rightarrow\)\(x< -\frac{4}{3};x\ne2\)

a) \(\frac{1}{\sqrt{x^2-8x+15}}\)DK : \(x^2-8x+15>0\Rightarrow x< 3\)hoặc \(x>5\)

b) \(\sqrt{2}-\sqrt{x-1}\)DK : \(x-1\ge0\Rightarrow x\ge1\)

help

f: ĐKXĐ: \(\dfrac{2x-1}{2-x}>=0\)

=>\(\dfrac{2x-1}{x-2}< =0\)

=>\(\dfrac{1}{2}< =x< 2\)

g: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>0\end{matrix}\right.\Leftrightarrow3< =x< 5\)

h: ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\x+5>=0\end{matrix}\right.\Leftrightarrow x>=1\)

biểu thức e viết liền quá khó phân biệt  ví dụ như x +1 -\(\frac{2\sqrt{x}}{\sqrt{x-1}}\)hay là x +\(\frac{1-\sqrt{2x}}{\sqrt{x-1}}\)

Điều kiện 4 – a ≥ 0 => -a ≥ -4 = > a ≤ 4