ĐK

\(\left\{{}\begin{matrix}x+3\ge0\\7-x\ge0\\2x-8\ge0\end{matrix}\right.\)

Giải hệ bất PT trên được ĐK tổng hợp là \(4\le x\le7\)

Bình phương 2 vế PT

\(x+3+7-x-2\sqrt{\left(x+3\right)\left(7-x\right)}=2x-8\)

\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(7-x\right)}=18-2x\)

BP 3 vế PT

\(4\left(x+3\right)\left(7-x\right)=324+4x^2-72x\)

\(\Leftrightarrow28x-4x^2+84-12x=324+4x^2-72x\)

\(\Leftrightarrow8x^2-88x+240=0\Leftrightarrow x^2-11x+30=0\)

Giải PT bậc 2 rồi đối chiếu với đk, bạn tự làm nốt nhé

Điều kiện xác định: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x-3\right)\left(\sqrt{x^2-4}-x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x^2-4}=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\\left\{{}\begin{matrix}x\ge-3\\x^2-4=x^2+6x+9\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\\left\{{}\begin{matrix}x\ge-3\\6x=-13\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\\left\{{}\begin{matrix}x\ge-3\\x=-\dfrac{13}{6}\end{matrix}\right.\end{matrix}\right.\)

Kết hợp với điều kiện xác định, ta được: \(\left[{}\begin{matrix}x=3\\x=-\dfrac{13}{6}\end{matrix}\right.\)

Vậy nghiệm của phương trình là S = \(\left\{-\dfrac{13}{6};3\right\}\)

ĐK

\(x\ge0\) và \(x+1\ge0\Leftrightarrow x\ge-1\)

\(\Rightarrow x\ge0\)

Bình phương 2 vế PT

\(x+x+1+2\sqrt{x\left(x+1\right)}=1+x\left(x+1\right)+2\sqrt{x\left(x+1\right)}\)

\(\Leftrightarrow2x+1=1+x^2+x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) Thỏa mãn điều kiện \(x\ge0\)

2\(\sqrt{x+2+\sqrt{x+1}}\) - \(\sqrt{x+1}\) = 4;  Đk \(x\ge\) -1

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

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

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

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

  \(\sqrt{x+1}\)       = 4 - 2

   \(\sqrt{x+1}\)       = 2

    \(x+1\)      = 4

    \(x\)              = 4 - 1

       \(x\)            = 3

\(...\Rightarrow2\sqrt[]{x+1+2\sqrt[]{x+1+1}}-\sqrt[]{x+1}=4\left(x\ge-1\right)\)

\(\Rightarrow2\sqrt[]{\left(\sqrt[]{x+1}+1\right)^2}-\sqrt[]{x+1}=4\)

\(\Rightarrow2|\sqrt[]{x+1}+1|-\sqrt[]{x+1}=4\left(1\right)\)

Nếu \(\sqrt[]{x+1}+1\ge0\Rightarrow x\ge-1\)

\(\left(1\right)\Rightarrow2\sqrt[]{x+1}+1-\sqrt[]{x+1}=4\)

\(\Rightarrow\sqrt[]{x+1}=3\Rightarrow x+1=9\Rightarrow x=8\)

Nếu \(\sqrt[]{x+1}+1\le0\Rightarrow x\in\varnothing\)

Vậy \(x=8\)

\(B=\left[\dfrac{\sqrt{x-2}}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\left[\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\dfrac{-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2x}{x-1}\)

b/

\(B=-\dfrac{2\left(x-1\right)+2}{x-1}=-2+\dfrac{2}{x-1}\)

Để B nguyên

\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)

Điều kiện

\(3x-1\ge0\Leftrightarrow x\ge\dfrac{1}{3}\) 

\(4-x\ge0\Leftrightarrow x\le4\)

Kết hợp 2 đk \(\Rightarrow\dfrac{1}{3}\le x\le4\)

Bình phương 2 vế PT

\(4\left(3x-1\right)=4-x\)

\(\Leftrightarrow12x-4=4-x\Leftrightarrow13x=8\)

\(\Leftrightarrow x=\dfrac{8}{13}\) Đối chiếu với đk thỏa mãn

   (\(\dfrac{1}{\sqrt{2}-1}\) - \(\dfrac{1}{\sqrt{2}+1}\)): \(\sqrt{3-2\sqrt{2}}\)

= \(\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)}\): \(\sqrt{2-2\sqrt{2}+1}\)

= \(\dfrac{2}{2-1}\).\(\sqrt{\left(\sqrt{2}-1\right)^2}\)

= 2(\(\sqrt{2}\)  - 1)

= 2\(\sqrt{2}\) - 2