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đặt 2x+3=a

\(y\sqrt{y}+y=a\sqrt{a}+a\)

=>\(\left(\sqrt{y}-\sqrt{a}\right)\left(y+\sqrt{ay}+a+\sqrt{a}+\sqrt{y}\right)=0\)

=>\(\sqrt{y}=\sqrt{a}\Rightarrow y=2x+3\)

thay vào Q tìm min là xong

Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\left(a>0\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}}\)

Thì ta có

\(\frac{b^2}{a^2}=\frac{a+1}{b+1}\)

\(\Leftrightarrow b^3+b^2=a^3+a^2\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2\right)+\left(b-a\right)\left(b+a\right)=0\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2+b+a\right)=0\)

Mà \(\left(b^2+ab+a^2+b+a\right)>0\)

\(\Rightarrow a=b\)

\(\Rightarrow2x+3=y\)

Thế vào Q ta được 

\(Q=2x^2-5x-12=\left(2x^2-\frac{2x\times\sqrt{2}\times5}{2\sqrt{2}}+\frac{25}{8}\right)-\frac{121}{8}\)

\(=\left(\sqrt{2}x-\frac{5}{2\sqrt{2}}\right)^2-\frac{121}{8}\ge\frac{-121}{8}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))

Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)

\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)

\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)

Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)

Phần đặt ẩn a,b,c bạn bỏ đi nhé ^^

\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)

\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)

\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)

\(=5\left(a+b\right)=5.2016=10080\)

alibaba nguyễn Em kiểm tra lại bài làm của mình nhé! 

Nguyễn Linh Chi haha, em nhìn ra rối, chỗ dấu "=" thứ 2 phải sửa lại thành dấu "+" ,còn anh ấy phân tích có sai chỗ nào thì em ko biết:D (hình như là đúng)