a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

c, ĐK: \(0\le x\le9\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(pt\Leftrightarrow9+2\sqrt{9x-x^2}=-x^2+9x+m\)

\(\Leftrightarrow-\left(-x^2+9x\right)+2\sqrt{9x-x^2}+9=m\)

\(\Leftrightarrow-t^2+2t+9=m\)

Khi \(m=9,pt\Leftrightarrow-t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9x-x^2=0\\9x-x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=9\left(tm\right)\\x=\dfrac{9\pm\sqrt{65}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình đã cho có nghiệm khi phương trình \(m=f\left(t\right)=-t^2+2t+9\) có nghiệm

\(\Leftrightarrow minf\left(t\right)\le m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le m\le10\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

Ui...... người ta nói nó dễ ..................................

\(2\sqrt{x+4}-4\sqrt{2x-6}=x-7\)

\(\Leftrightarrow\sqrt{2^2\left(x+4\right)}-\sqrt{4^2\left(2x-6\right)}=x-7\)

\(\Leftrightarrow\sqrt{4x+16}-\sqrt{32x-96}=x-7\)

\(\Leftrightarrow\left(\sqrt{4x+16}-\sqrt{32x-96}\right)^2=\left(x-7\right)^2\)

\(\Leftrightarrow\sqrt{4x+16}^2-2.\sqrt{4x+16}.\sqrt{32x-96}+\sqrt{32x-96}^2=x^2-14x+49\)

\(\Leftrightarrow\left(4x+16\right)-2.\sqrt{\left(4x+16\right)\left(32x-96\right)}+\left(32x-96\right)=x^2-14x+49\)

\(\Leftrightarrow\left(4x+16\right)-2.\sqrt{128x^2-384x+512x-1536}+\left(32x-96\right)=x^2-14x+49\)

\(\Leftrightarrow\left(-2\sqrt{128x^2-384x+512x-1536}\right)=\left[x^2-14x+49-\left(4x+16\right)-\left(32x-96\right)\right]\)

\(\Leftrightarrow\left(-2\sqrt{128x^2+128x-1536}\right)^2=\left(x^2-50x+129\right)^2\)

\(\Leftrightarrow4.\left(128x^2+128x-1536\right)=\left(x^2-50x\right)^2+2.\left(x^2-50x\right).129+129^2\)

\(\Leftrightarrow512x^2+512x-6144=\left(x^2-50x\right)^2+258.\left(x^2-50x\right)+16641\)

\(\Leftrightarrow512x^2+512x-6144=x^4-100x^3+2500x^2+258x^2-12900x+16641\)

\(\Leftrightarrow-x^4+100x^3-2246x^2+13412x-22785=0\)

\(\Leftrightarrow x_1\approx70,94\) ; \(x_2\approx3,0588\) ; \(x_3=21\) ; \(x_4=5\)

Bài này có 1 nghiệm duy nhất thôi nha : x = 5 

tại máy tính của mình ra sai kết quả 

ĐKXĐ: $x \geq 2$

\(\Leftrightarrow2\left(x-4\right).\sqrt{x-2}-2\left(x-4\right)+\left(x-2\right)\sqrt{x+1}-2\left(x-2\right)+6x-18=0\\ \Leftrightarrow2.\left(x-4\right).\dfrac{x-3}{\sqrt{x-2}+1}+\left(x-2\right).\dfrac{x-3}{\sqrt{x+1}+2}+6.\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=0\right)\\ \Leftrightarrow x=3\)

Vì \(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=\dfrac{2\left(x-4\right)+4.\sqrt{x-2}+4}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2\\ =\dfrac{2\left(x-2\right)+4.\sqrt{x-2}}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2>0\)

Vậy....