Mn giúp e với ak e đag cần gấp
\(1,\)
\(a,M=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\left(x\ge4\right)\\ \Leftrightarrow M^2=2x+2\sqrt{\left(x+4\sqrt{x-4}\right)\left(x-4\sqrt{x-4}\right)}\\ \Leftrightarrow M^2=2x+2\sqrt{x^2-16x+64}\\ \Leftrightarrow M^2=2x+2\sqrt{\left(x-8\right)^2}\\ \Leftrightarrow M^2=2x+2\left(x-8\right)=2x+2x-8=4x-16=4\left(x-4\right)\\ \Leftrightarrow M=2\sqrt{x-4}\)
\(b,\)Chịu
\(2,\)
\(a,A=\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(x-3\sqrt{x}+2\right)\left(x\ge0;x\ne4\right)\\ \Leftrightarrow A=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\\ \Leftrightarrow A=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,\)Để \(A< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{1}{2}< 0\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}}{2\sqrt{x}}< 0\Leftrightarrow\dfrac{2\sqrt{x}-2}{2\sqrt{x}}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{x}-2< 0\\2\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{x}-2>0\\2\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 1\\x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow0< x< 1\)
Vậy \(0< x< 1\) thì \(A< \dfrac{1}{2}\)
\(c,\)Để \(A\in Z\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}\in Z\Leftrightarrow1-\dfrac{1}{\sqrt{x}}\in Z\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}\inƯ\left(1\right)=\left\{-1;1\right\}\)
\(\Leftrightarrow x\in\left\{1\right\}\)
Vậy \(x=1\) thì \(A\in Z\)
\(3,\)
\(a,\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(x>0,x\ne9\right)\\ =\dfrac{x+3+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(b,x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ \Leftrightarrow x=5+\sqrt{2}-4-\sqrt{2}=1\)
Thay vào \(B\), ta được: \(B=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(c,\) Ta có \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
Mà \(x>0\)
\(\Leftrightarrow\sqrt{x}>0\\ \Leftrightarrow\sqrt{x}+3>3\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}< \dfrac{2}{3}\\ \Leftrightarrow1-\dfrac{2}{\sqrt{x}+3}>1-\dfrac{2}{3}=\dfrac{1}{3}\)
Tick full hộ nha 🤩
