A có nghĩa khi mẫu phân số khác 0 và biểu thức trong căn dương;

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}^2-\sqrt{x}+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\\\forall x\end{cases}}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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bạn chỉ cần cố gắng là làm được

Vì \(0\le x\le\frac{1}{2}\Rightarrow0\le2x\le1\Rightarrow-1\le-2x\le0\)

\(1-1\le1-2x\le1\Rightarrow0\le x\left(1-2x\right)\le\frac{1}{2}\)

\(0\le M\le\frac{1}{2}\Rightarrow M_{max}=\frac{1}{2}\)

Điều kiện để hệ có nghiệm duy nhất là \(\frac{1}{a}\ne\frac{-1}{1}\Leftrightarrow a\ne-1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engelta có:

\(VT=\frac{700}{2\left(xy+yz+xz\right)}+\frac{386}{x^2+y^2+z^2}\)\(=\frac{\sqrt{700}^2}{2\left(xy+yz+xz\right)}+\frac{\sqrt{386}^2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{700}+\sqrt{386}\right)^2}{x^2+y^2+z^2+2\left(xy+yz+xz\right)}\)\(=\frac{\left(\sqrt{700}+\sqrt{386}\right)^2}{\left(x+y+z\right)^2}\)

\(=\left(\sqrt{700}+\sqrt{386}\right)^2>2015\left(x+y+z=1\right)\)

\(=\frac{\left(x^2+5x+6\right)+x\sqrt{9-x^2}}{\left(3x-x^2\right)+\left(2+x\right)\sqrt{9-x^2}}\)

\(=\frac{\left(x+2\right)\left(3+x\right)+x\sqrt{\left(3+x\right)\left(3-x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3+x\right)\left(3-x\right)}}\) nhóm nhân tử chung

\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{3+x}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{3+x}\right)}\)rồi rút gọn được

\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)

ai tích giùm tôi không 1 tích thôi cũng không ddc sao