thánh

em mới học lớp 5 

\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)

\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)

\(P_{min}=6\sqrt{5}\) khi \(x=y=z=1\)

Mặt khác với mọi \(x\in\left[0;3\right]\) ta có:

\(\sqrt{x^2-6x+25}\le\dfrac{15-x}{3}\)

Thật vậy, BĐT tương đương: \(9\left(x^2-6x+25\right)\le\left(15-x\right)^2\)

\(\Leftrightarrow8x\left(3-x\right)\ge0\) luôn đúng

Tương tự: ...

\(\Rightarrow P\le\dfrac{45-\left(x+y+z\right)}{3}=14\)

\(P_{max}=14\) khi \(\left(x;y;z\right)=\left(0;0;3\right)\) và hoán vị

sol của tớ :3

Nếu y=0 thì x²=1 => P=2

Nếu y\(\ne\)0 .Đặt \(t=\frac{x}{y}\)

\(P=\frac{2\left(x^2+6xy\right)}{1+2xy+2y^2}=\frac{2\left(x^2+6xy\right)}{x^2+2xy+3y^2}=\frac{2\left[\left(\frac{x}{y}\right)^2+6\cdot\frac{x}{y}\right]}{\left(\frac{x}{y}\right)^2+2\frac{x}{y}+3}=\frac{2\left(t^2+6t\right)}{t^2+2t+3}\)

\(\Rightarrow P.t^2+2P\cdot t+3P=2t^2+12t\)

\(\Leftrightarrow t^2\left(P-2\right)+2t\left(P-6\right)+3P=0\)

Xét \(\Delta'=\left(P-2\right)^2-3P\left(P-6\right)=-2P^2-6P+36\ge0\)

\(\Leftrightarrow-6\le P\le3\)

Dấu bằng xảy ra khi:

Max:\(x=\frac{3}{\sqrt{10}};y=\frac{1}{\sqrt{10}}\left(h\right)x=\frac{3}{-\sqrt{10}};y=\frac{1}{-\sqrt{10}}\)

Min:\(x=\frac{3}{\sqrt{13}};y=-\frac{2}{\sqrt{13}}\left(h\right)x=-\frac{3}{\sqrt{13}};y=\frac{2}{\sqrt{13}}\)

khó ha

Ta co:

\(P=\frac{2x^2+12xy}{1+2xy+2y^2}\)

\(\Leftrightarrow Px^2+Py^2+2Pxy+2Py^2=2x^2+12xy\)

\(\Leftrightarrow\left(P-2\right)x^2+\left(2P-12\right)xy+3Py^2=0\)

\(\Leftrightarrow\left(P-2\right)\frac{x^2}{y^2}+\left(2P-12\right)\frac{x}{y}+3P=0\)

Dat \(\frac{x}{y}=t\left(t\in R\right)\)

PT tro thanh

\(\left(P-2\right)t^2+\left(2P-12\right)t+2P=0\)

Xet \(P=2\)\(\Rightarrow x=\frac{5}{4};y=\frac{5}{3}\)

Xet \(P\ne2\)

Ta lai co:

\(\Delta^`\ge0\)

\(\Leftrightarrow\left(P-6\right)^2-\left(P-2\right).2P\ge0\)

\(\Leftrightarrow-P^2-8P+36\ge0\)

\(\Leftrightarrow P^2+8P-36\le0\)

\(\Leftrightarrow\left(P+4-2\sqrt{13}\right)\left(P+4+2\sqrt{13}\right)\le0\)

TH1:

\(\hept{\begin{cases}P+4-2\sqrt{13}\ge0\\P+4+2\sqrt{13}\le0\end{cases}\left(l\right)}\)

TH2:

\(\hept{\begin{cases}P+4-2\sqrt{13}\le0\\P+4+2\sqrt{13}\ge0\end{cases}\Leftrightarrow-4-2\sqrt{13}\le P\le2\sqrt{13}-4}\)

Dau '=' xay ra khi \(\frac{x}{y}=\frac{10-2\sqrt{13}}{2\sqrt{13}-6}\Leftrightarrow\Leftrightarrow\hept{\begin{cases}x=\frac{1}{\sqrt{1+\left(\frac{\text{ }2\sqrt{13}-6}{10-2\sqrt{13}}\right)^2}}\\y=\frac{2\sqrt{13}-6}{\left(10-2\sqrt{13}\right)\sqrt{1+\left(\frac{2\sqrt{13}-6}{10-2\sqrt{13}}\right)^2}}\end{cases}}\)

Cho dau '=' xay ra khung chac dung khong nua

\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

đề bài sai r bn ơi phải là +10 chứ ko phải +8 đâu nhá