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ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-2\\y\ge0\\y\ne1\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\dfrac{2}{\sqrt{y}-1}=4\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\sqrt{y}=\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{9}{4}\left(tm\right)\end{matrix}\right.\)

\(\left(a^2+1\right)\left(2x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}a^2+1=0\\2x+8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}a^2=-1\left(vô.lí\right)\\2x=-8\end{matrix}\right.\\ \Rightarrow x=-4\)

\(ĐKXĐ:\left\{{}\begin{matrix}1-x\ne0\\1+x\ne0\\1-x^2\ne0\\x^2-2x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne\pm1\\\left(x-1\right)^2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x\ne1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)

\(A=\left(\dfrac{1+x}{1-x}-\dfrac{1-x}{1+x}+\dfrac{4x^2}{1-x^2}\right):\dfrac{4x^2-4}{x^2-2x+1}\)

\(\Leftrightarrow A=\dfrac{\left(1+x\right)^2-\left(1-x\right)^2+4x^2}{\left(1-x\right)\left(1+x\right)}:\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

\(\Leftrightarrow A=\dfrac{1+2x+x^2-1+2x-x^2+4x^2}{\left(1-x\right)\left(1+x\right)}:\dfrac{4\left(x+1\right)}{x-1}\)

\(\Leftrightarrow A=\dfrac{4x^2+4x}{\left(1-x\right)\left(1+x\right)}.\dfrac{x-1}{4\left(x+1\right)}\)

\(\Leftrightarrow A=\dfrac{-4x\left(x+1\right)}{\left(x-1\right)\left(1+x\right)}.\dfrac{x-1}{4\left(x+1\right)}\)

\(\Leftrightarrow A=\dfrac{-x}{1+x}\)

\(A=x+\dfrac{1}{x-2}\\ \Rightarrow A=x-2+\dfrac{1}{x-2}+2\)

Áp dụng BĐT Cô-si ta có:

\(A=x-2+\dfrac{1}{x-2}+2\\ \ge2\sqrt{\left(x-2\right).\dfrac{1}{x-2}}+2\\ =2\sqrt{1}+2\\ =4\)

 \(\text{Dấu "=" xảy ra}\Leftrightarrow x-2=\dfrac{1}{x-2}\\ \Leftrightarrow\left(x-2\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy \(A_{min}=4\Leftrightarrow x=3\)

a, thay m=2 vào hệ ta có:

\(\left\{{}\begin{matrix}2x+y=2+1\\4x+2y=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x+y=3\\2x+y=1\left(vô.lí\right)\end{matrix}\right.\)

Ta có:MB=AB-AM=24-16=8

Áp dụng định lý Ta-lét ta có:

\(\dfrac{AM}{MB}=\dfrac{AN}{NC}\\ \Rightarrow\dfrac{16}{8}=\dfrac{14}{x}\\ \Rightarrow2=\dfrac{14}{x}\\ \Rightarrow x=7\)