1.

đk: \(x\ge2\)

Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:

ta có : \(x^3-3x^2+2y^3-6x=0\)

\(\Leftrightarrow x^3-3xy^2+2y^3=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)

ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)

\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)

\(pt\Leftrightarrow x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)

\(\Leftrightarrow x^3-x\left(x+2\right)+2\sqrt{\left(x+3\right)^2}-2x\left(x+2\right)=0\)

\(\Leftrightarrow x\left[x^2-\left(x+2\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(\sqrt{x+2}-x\right)^2\left(2\sqrt{x+2}+x\right)=0\)

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

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)

\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(x=y\Rightarrow M=x^{11}-x^{2018}\)

\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)

ĐKXĐ:...

\(\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x-2018}-\sqrt{y-2018}+x^2-y^2=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(x+y\right)}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\frac{x-y}{\sqrt{x-2018}+\sqrt{y-2018}}+\left(x-y\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{x+y}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\frac{1}{\sqrt{x-2018}+\sqrt{y-2018}}+x+y\right)=0\)

\(\Leftrightarrow x=y\) (ngoặc phía sau luôn dương)

Thay vào M chẳng được cái gì cả, \(M=x^{11}-x^{2018}\) :(

Chắc bạn nhầm đề

Cách của mình dài ,bạn nào có cách khác ngắn gọn hơn thì chỉ cho mình với ạ. Cảm ơn

Trước hết ta chứng minh  BĐT phụ sau: \(\sqrt{a^2+b^2}+\sqrt{x^2+y^2}\ge\sqrt{\left(a+x\right)^2+\left(b+y\right)^2}.\)(*)

Thật vậy: \(ax+by\le\sqrt{\left(ax+by\right)^2}\le\sqrt{\left(a^2+b^2\right)\left(x^2+y^2\right)}\)(BĐT bunhiacopxi)

\(\Leftrightarrow a^2+b^2+x^2+y^2+2\sqrt{\left(a^2+b^2\right)\left(x^2+y^2\right)}\ge a^2+b^2+x^2+y^2+2\left(ax+by\right)\)

\(\Leftrightarrow\left(\sqrt{a^2+b^2}+\sqrt{x^2+y^2}\right)^2\ge\left(a+x\right)^2+\left(b+y\right)^2\)

\(\Leftrightarrow\sqrt{a^2+b^2}+\sqrt{x^2+y^2}\ge\sqrt{\left(a^2+b^2\right)\left(x^2+y^2\right)}\). BĐT đã được chứng minh

Xét : \(\left(x+\sqrt{1+x^2}\right)\left(x-\sqrt{1+x^2}\right)=x^2-\left(1+x^2\right)=-1.\)

Theo giả thết : \(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=2018\)

\(\Rightarrow2018\left(x-\sqrt{1+x^2}\right)=-\left(y+\sqrt{1+y^2}\right).\)

\(\Leftrightarrow2018x+y=2018\sqrt{1+x^2}-\sqrt{1+y^2}.\)(1)

Tương tự:

Xét:\(\left(y+\sqrt{1+y^2}\right)\left(y-\sqrt{1+y^2}\right)=y^2-\left(1+y^2\right)=-1\)

Theo giả thiết : \(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=2018\)

\(\Rightarrow2018\left(y-\sqrt{1+y^2}\right)=-\left(x+\sqrt{1+x^2}\right)\)

\(\Leftrightarrow x+2018y=-\sqrt{1+x^2}+2018\sqrt{1+y^2}\)(2)

Cộng các vế của (1) và (2) lại ta được

\(2019\left(x+y\right)=2017\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\)

Khi đó áp dụng bất đẳng thức (*) ta có;

\(2019\left(x+y\right)=2017\left(\sqrt{1^2+x^2}+\sqrt{1^2+y^2}\right)\ge2017\left(\sqrt{\left(1+1\right)^2+\left(x+y\right)^2}\right)\)

\(\Rightarrow2019\left(x+y\right)\ge2017\sqrt{4+\left(x+y\right)^2}\)

Đặt \(x+y=a>0\)ta có;

\(2019a\ge2017\sqrt{4+a^2}\Leftrightarrow2019^2a^2\ge2017^2a^2+2017^2.4\)

\(\Leftrightarrow\left(2019^2-2017^2\right)a^2\ge\left(2017.2\right)^2\Leftrightarrow a^2\ge\frac{2017^2.2.2}{2.4036}\Leftrightarrow a^2\ge\frac{2017^2}{2018}\)

\(\Rightarrow a\ge\frac{2017}{\sqrt{2018}}\Rightarrow x+y\ge\frac{2017}{\sqrt{2018}}.\)

Vậy giá trị nhỏ nhất của biểu thức P=x+y là \(\frac{2017}{\sqrt{2018}}\)

Dấu '=' xảy ra khi \(x=y=\frac{2017}{2\sqrt{2018}}.\)

bn đào thu hà k cần cm bdt phụ đâu đấy là bdt mincopski đc dùng luôn

P/s : làm bừa thôi!

\(\sqrt{x-2018}+\sqrt{x^2+11}+x^2=\sqrt{y^2+11}+\sqrt{y-2018}+y^2\)

\(\Leftrightarrow x=y\)

\(\Rightarrow M=x^{11}-x^{2018}\)

Đến đây em tịt !!

ĐKXĐ: \(\left\{{}\begin{matrix}2020-y^2\ge0\\2020-z^2\ge0\\2020-x^2\ge0\end{matrix}\right.\)

Ta có:

\(x\sqrt{2020-y^2}+y\sqrt{2020-z^2}+z\sqrt{2020-x^2}=3030\)

\(\Leftrightarrow2x\sqrt{2020-y^2}+2y\sqrt{2020-z^2}+2z\sqrt{2020-x^2}=6060\)

\(\Leftrightarrow2020-y^2-2x\sqrt{2020-y^2}+x^2+2020-z^2-2y\sqrt{2020-z^2}+y^2+2020-x^2-2z\sqrt{2020-x^2}+z^2=0\)

   \(\Leftrightarrow\left(\sqrt{2020-y^2}-x\right)^2+\left(\sqrt{2020-z^2}-y\right)^2+\left(\sqrt{2020-x^2}-z\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{2020-y^2}-x\right)^2=\left(\sqrt{2020-z^2}-y\right)^2=\left(\sqrt{2020-x^2}-z\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2020-y^2}=x\\\sqrt{2020-z^2}=y\\\sqrt{2020-x^2}=z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2020-y^2=x^2\\2020-z^2=y^2\\2020-x^2=z^2\end{matrix}\right.\)(vì \(x,y,z>0\))

\(\Leftrightarrow\left\{{}\begin{matrix}2020=x^2+y^2\\2020=y^2+z^2\\2020=z^2+x^2\end{matrix}\right.\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)=3.2020\)

\(\Rightarrow x^2+y^2+z^2=3.1010=3030\)

\(\Rightarrow A=x^2+y^2+z^2=3030\)

Vậy \(A=3030\)