Lời giải:

Đặt $\frac{y}{x}=a(a>0)$ thì:

\(P=\sqrt{\frac{1}{1+(\frac{2y}{x})^3}}+\sqrt{\frac{4}{1+(1+\frac{x}{y})^3}}=\sqrt{\frac{1}{1+8a^3}}+\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\)

Áp dụng BĐT AM-GM dạng $xy\leq \left(\frac{x+y}{2}\right)^2$ ta có:

\(1+8a^3=1+(2a)^3=(1+2a)(1-2a+4a^2)\leq \left(\frac{1+2a+1-2a+4a^2}{2}\right)^2=(2a^2+1)^2\)

\(\Rightarrow \sqrt{\frac{1}{8a^3+1}}\geq \frac{1}{2a^2+1}(1)\)

\(1+(1+\frac{1}{a})^3=(2+\frac{1}{a})[1-(1+\frac{1}{a})+(1+\frac{1}{a})^2]\leq (\frac{3a^2+2a+1}{2a^2})^2\)

\(\Rightarrow \sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{4a^2}{3a^2+2a+1}\)

Mà: \(\frac{4a^2}{3a^2+2a+1}\geq \frac{4a^2}{3a^2+a^2+1+1}=\frac{2a^2}{2a^2+1}\) nên \(\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{2a^2}{2a^2+1}(2)\)

Từ $(1);(2)\Rightarrow P\geq \frac{1}{2a^2+1}+\frac{2a^2}{2a^2+1}=1$

Vậy $P_{\min}=1$ khi $a=1\Leftrightarrow x=y$

Ta có :

\(A=\sqrt{\frac{x^3}{x^3+8y^3}}\)

\(\Rightarrow A=\sqrt{\frac{1}{1+\left(\frac{2y}{x}\right)^3}}\)

\(\Rightarrow A=\sqrt{\frac{1}{\left(1+\frac{2y}{x}\right)\left(1-\frac{2y}{x}+\frac{4y^2}{x^2}\right)}}\)

\(\Rightarrow A\ge\frac{1}{\frac{\left(1+\frac{2y}{x}\right)+\left(1-\frac{2y}{x}+\frac{4y^2}{x^2}\right)}{2}}\)

\(\Rightarrow A\ge\frac{2}{2+\frac{4y^2}{x^2}}=\frac{1}{1+2\left(\frac{y}{x}\right)^2}\)

VÀ

\(B=\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(\Rightarrow B=\sqrt{\frac{4}{1+\left(\frac{x}{y}+1\right)^3}}\)

\(\Rightarrow B=\frac{2}{\sqrt{\left[1+\left(1+\frac{x}{y}\right)\right]\left[1-\left(1+\frac{x}{y}\right)+\left(1+\frac{x}{y}\right)^2\right]}}\)

\(\Rightarrow B\ge\frac{2}{\frac{\left[1+\left(1+\frac{x}{y}\right)\right]+\left[1-\left(1+\frac{x}{y}\right)+\left(1+\frac{x}{y}\right)^2\right]}{2}}\)

\(\Rightarrow B\ge\frac{4}{2+\left(1+\frac{x}{y}\right)^2}\)

Suy ra :

\(P=A+B\ge\frac{1}{1+2\left(\frac{y}{x}\right)^2}+\frac{4}{2+\left(1+\frac{x}{y}\right)^2}\)

\(\Rightarrow P\ge\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\)

\(\Rightarrow P\ge\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2y^2+2\left(x^2+y^2\right)}=\frac{x^2}{x^2+2y^2}+\frac{4y^2}{2x^2+4y^2}=\frac{x^2}{x^2+2y^2}+\frac{2y^2}{x^2+2y^2}=1\)

"=" khi \(x=y\)

lô bn xàm lồn

bn trẩu , m phải ARMY hơm 

nếu phải thì nhục quá trời,tự nhiên fan BTS lại chưa con phò như mài ,u hú hú

bớt sàm lại đuy,ko thì đừng làm AMI nx,ư~~

\(Q=\frac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\frac{2y^2}{\sqrt{y\left[y^3+\left(x+y\right)^3\right]}}\)

\(=\frac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\frac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}\)

\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\)\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)

\(\Rightarrow Q\ge1\).Vậy Min_Q=1

\(Q=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^4+8xy^3=x^4+8.xy.y^2\le x^4+4\left(x^2y^2+y^4\right)=\left(x^2+2y^2\right)^2\)

\(\Rightarrow\frac{x^2}{\sqrt{x^3+8xy^3}}\ge\frac{x^2}{x^2+2y^2}\)

\(\sqrt{y\left(y^3+\left(x+y\right)^3\right)}=\sqrt{\left(xy+2y^2\right)\left(x^2+y^2+xy\right)}\le\frac{x^2+3y^2+2xy}{2}=\frac{2y^2+\left(x+y\right)^2}{2}\)

\(\le\frac{2y^2+2\left(x^2+y^2\right)}{2}=x^2+2y^2\)

\(\Rightarrow Q\ge\frac{x^2}{x^2+2y^2}+\frac{2y^2}{x^2+2y^2}=1\)

Vậy minQ= 1 tại \(x=y>0\)

Bonus cho các bạn 1 cách rất hay nè:

\(Q=\sqrt{\frac{1}{1+8\left(\frac{y}{x}\right)^3}}+\sqrt{\frac{4}{1+\left(\frac{x}{y}+1\right)^3}}\)        (CHIA CẢ TỬ VÀ MẪU LẦN LƯỢT CHO \(x^3;y^3\))

Đặt: \(\frac{y}{x}=a;\frac{x}{y}=b\)   => ab=1

=>    \(Q=\sqrt{\frac{1}{1+8a^3}}+\sqrt{\frac{4}{1+\left(b+1\right)^3}}\)

=>    \(Q=\frac{2}{2\sqrt{\left(2a+1\right)}.\sqrt{4a^2-2a+1}}+\frac{4}{2\sqrt{\left(b+2\right)}.\sqrt{b^2+b+1}}\)

=>    \(Q\ge\frac{2}{4a^2+2}+\frac{4}{b^2+2b+3}\)

=>    \(Q\ge\frac{1}{2a^2+1}+\frac{4}{b^2+3+b^2+1}\)

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

Ta thay lại 2 giá trị ẩn phụ đã đặt ở trên đó là: \(a=\frac{y}{x};b=\frac{x}{y}\)

Khi đó       \(Q\ge\frac{1}{2\left(\frac{y}{x}\right)^2+1}+\frac{2}{\left(\frac{x}{y}\right)^2+2}\)

=>    \(Q\ge\frac{1}{\frac{2y^2+x^2}{x^2}}+\frac{2}{\frac{x^2+2y^2}{y^2}}\)

=>     \(Q\ge\frac{x^2}{x^2+2y^2}+\frac{2y^2}{x^2+2y^2}=\frac{x^2+2y^2}{x^2+2y^2}=1\)

Vậy Q min = 1 <=> \(x=y\)

Lời giải:
\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\)

Xét:

\(x^4+8xy^3-(x^2+2y^2)^2=8xy^3-4y^4-4x^2y^2\)

\(=-4y^2(x^2-2xy+y^2)=-4y^2(x-y)^2\leq 0\)

\(\Rightarrow x^4+8xy^3\leq (x^2+2y^2)^2\)

\(\Rightarrow \frac{x^2}{\sqrt{x^4+8xy^3}}\geq \frac{x^2}{x^2+2y^2}(*)\)

Mặt khác:
\(y^4+y(x+y)^3-(x^2+2y^2)^2=x^3y+3xy^3-2y^4-x^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^2(y-x)(y+x)\)

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

\(=(y-x)[(x-y)(x^2+xy+y^2)+y^2(x-y)]\)

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

\(\Rightarrow y^4+y(x+y)^3\leq (x^2+2y^2)^2\Rightarrow \frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\geq \frac{2y^2}{x^2+2y^2}(**)\)

Từ $(*); (**)\Rightarrow A\geq 1$

\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)

Ta xét các trường hợp sau:

Trường hợp 1:

\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:

\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)

\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)

Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)

Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:

\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)

Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)

+ Nếu như:

\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)

+ Nếu như:

\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)

Trường hợp 2:

\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:

\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)

Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)

Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v

3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)

\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)

Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)

\(\Leftrightarrow2b=1-\frac{1}{a}\)

Thay vào (1) ta được :

\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)

Giải pt được \(a=1\)

Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)

Ta có hệ :

\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy...