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1.
\(\Delta=\left(m+2\right)^2-4m=m^2+4>0;\forall m\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=m\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{x_1+x_2-2}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{x_1+x_2-2}\)
\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m+2-2}\)
\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m}\)
\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m+2=1\end{matrix}\right.\) \(\Rightarrow m=-1\) (thỏa)
2.
\(\Delta'=\left(m-2\right)^2-2\left(2m-6\right)=\left(m-4\right)^2\)
Pt có 2 nghiệm pb khi \(\left(m-4\right)^2>0\Rightarrow m\ne4\)
Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m+2\\x_1x_2=m-3\end{matrix}\right.\)
\(2x_1x_2-\left(x_1-x_2\right)^2=-1\)
\(\Leftrightarrow2x_1x_2-\left(x_1+x_2\right)^2+4x_1x_2=-1\)
\(\Leftrightarrow6\left(m-3\right)-\left(-m+2\right)^2=-1\)
\(\Leftrightarrow-m^2+10m-21=0\Rightarrow\left[{}\begin{matrix}m=3\\m=7\end{matrix}\right.\) (thỏa mãn)
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\(D=\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\left(\sqrt{x}-3\right)+7}{\sqrt{x}-3}=2+\dfrac{7}{\sqrt{x}-3}\left(x\ge0;x\ne9\right)\)
Để \(D\) nguyên thì \(2+\dfrac{7}{\sqrt{x}-3}\) nguyên
\(\Rightarrow \dfrac{7}{\sqrt x-3}\) nguyên
\(\Rightarrow7⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(7\right)\)
\(\Rightarrow\sqrt{x}-3\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{4;10;2;-3\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{4;10;2\right\}\)
\(\Rightarrow x\in\left\{16;100;4\right\}\left(tm.đk.x.nguyên\right)\)
Kết hợp với điều kiện, ta được: \(x\in\left\{4;16;100\right\}\)
\(D=\dfrac{2\sqrt[]{x}+1}{\sqrt[]{x}-3}\in Z\left(x\ge0;x\ne9\right)\)
\(\Leftrightarrow2\sqrt[]{x}+1⋮\sqrt[]{x}-3\)
\(\Leftrightarrow2\sqrt[]{x}+1-2\left(\sqrt[]{x}-3\right)⋮\sqrt[]{x}-3\)
\(\Leftrightarrow2\sqrt[]{x}+1-2\sqrt[]{x}+6⋮\sqrt[]{x}-3\)
\(\Leftrightarrow7⋮\sqrt[]{x}-3\)
\(\Leftrightarrow\sqrt[]{x}-3\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)
\(\Leftrightarrow x\in\left\{4;16;100\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-Đặt ĐK: x>-1;
-Đặt a=\(\sqrt{x+1}\);b=\(\sqrt{x^2-x+1}\); Ta được: 5ab=2(a2+b2)
-Phân tích thành nhân tử được :(a-2b)(2a-b)=0
Đến đây bạn giải tiếp đi :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x=0\Leftrightarrow\left\{{}\begin{matrix}5-m=y\\3+m=y\end{matrix}\right.\Leftrightarrow y=4\Leftrightarrow5-m=4\Leftrightarrow m=1\)
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}=\dfrac{x\sqrt{2}}{\sqrt{2x}\left(\sqrt{2}+\sqrt{x}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}=\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}=1\)