\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

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

           \(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)

\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)

\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-2>0\\x-4\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x>2\\x\ne4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

mik thấy đề sai sai

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giúp mình với ạ

a:Sửa đề: \(A=\left(\frac{\sqrt{x}-2}{x\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\cdot\left(\frac{1-x}{2}\right)\)

Thay x=4 vào A, ta được:

\(A=\left(\frac{\sqrt4-2}{4\cdot\sqrt4-1}-\frac{\sqrt4+2}{4+\sqrt4+1}\right)\cdot\frac{1-4}{2}=\left(-\frac{4}{4+2+1}\right)\cdot\frac{-3}{2}=\frac47\cdot\frac32=\frac{2\cdot3}{7}=\frac67\)

b: \(A=\left(\frac{\sqrt{x}-2}{x\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\cdot\left(\frac{1-x}{2}\right)\)

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

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

\(=\frac{\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(x+\sqrt{x}+1\right)}\cdot\frac{-\left(\sqrt{x}+1\right)}{2}=\frac{-x}{\left(x+\sqrt{x}+1\right)\cdot2}\cdot\frac{-\left(\sqrt{x}+1\right)}{1}=\frac{x\left(\sqrt{x}+1\right)}{2\left(x+\sqrt{x}+1\right)}\)

a: \(A=\dfrac{x+2\sqrt{x}+x-3\sqrt{x}+2-x-\sqrt{x}-2}{x-4}\)

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

ĐKXĐ: \(x\ge0;x\ne1\)

\(P=\dfrac{3x-2\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(2\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3x-2\sqrt{x}-4-x+1-2x-6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{-8\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Đề bài có vẻ không hợp lý