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\(P=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{2x+3\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+1+10\left(\sqrt{x}+1\right)-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+1+10\sqrt{x}+10-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{6}{\sqrt{x}+1}\)
b) Để P nguyên tố thì \(\frac{6}{\sqrt{x}+1}\) nguyên tố
Để \(P\inℕ^∗\) thì \(\sqrt{x}+1\inƯ\left(6\right)\)
Mà P nguyên tố \(\Rightarrow\frac{6}{\sqrt{x}+1}=\left\{2;3\right\}\Rightarrow\sqrt{x}+1=\left\{2;3\right\}\)
Với \(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với \(\sqrt{x}+1=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy ...........
Từ điều kiện suy ra \(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)
Áp dụng BĐT Cô-si, ta có :
\(3\le\sqrt{xy}+\sqrt{x}.1+\sqrt{y}.1\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}\)
\(\Rightarrow x+y\ge2\)
Ta có : \(\frac{x^2}{y}+y\ge2\sqrt{\frac{x^2}{y}.y}=2x\); \(\frac{y^2}{x}+x\ge2\sqrt{\frac{y^2}{x}.x}=2y\)
\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}+x+y\ge2x+2y\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge2\)
Vậy GTNN của P là 2 khi x = y = 1
Có: \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=\sqrt{2019}\)
\(\Leftrightarrow\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow\left[y\left(1+x^2\right)+x\left(1+y^2\right)\right]^2=2018\)
\(\Leftrightarrow y\left(1+x^2\right)+x\left(1+y^2\right)=\sqrt{2018}\)
hay \(A=\sqrt{2018}\)
Ta có:
\(1.\sqrt{1+x^2}+1.\sqrt{2x}\le\sqrt{\left(1+1\right)\left(1+x^2+2x\right)}=\sqrt{2}\left(x+1\right)\)
Tương tự:
\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\) ; \(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
Cộng vế:
\(P\le\sqrt{2}\left(x+y+z+3\right)+\left(2-\sqrt{2}\right)\left(x+y+z\right)\le\sqrt{2}\left(3+3\right)+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)
\(P_{max}=6+3\sqrt{2}\) khi \(x=y=z=1\)
\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)
\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)
\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)
\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)
\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)
\(\Rightarrow A\ge2\)
\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
Làm tiếp ạ
\(\Rightarrow P\ge\frac{289}{16}\)
Dấu"="Xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
Vậy MIN P=\(\frac{289}{16}\)\(\Leftrightarrow x=y=\frac{1}{2}\)
Em chả có cách gì ngoài cô si mù mịt :v
\(\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)\)
\(=\left(x^2+\frac{1}{16y^2}+\frac{1}{16y^2}+.....+\frac{1}{16y^2}\right)\left(y^2+\frac{1}{16x^2}+\frac{1}{16x^2}+.....+\frac{1}{16x^2}\right)\)
\(\ge17\sqrt[17]{\frac{x^2}{16^{16}\cdot y^{32}}}\cdot17\sqrt[17]{\frac{y^2}{16^{16}\cdot x^{32}}}\)
\(=17^2\sqrt[17]{\frac{x^2y^2}{16^{32}\cdot x^{32}\cdot y^{32}}}\)
\(=17^2\sqrt[17]{\frac{1}{16^{32}\cdot\left(xy\right)^{30}}}\)
\(\ge17^2\sqrt[17]{\frac{1}{16^{32}\left(\frac{x+y}{2}\right)^{60}}}=\frac{289}{16}\)
Dấu "=" xảy ra tại x=y=1/2
Ta có: \(\sqrt{x+1}+\sqrt{y-1}\le\sqrt{2\left(x+y\right)}\)
\(\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+8}\le\sqrt{2\left(x+y\right)}\)
\(\Leftrightarrow2\left(x-y\right)+10x-6y+8\le2\left(x+y\right)\)
\(\Leftrightarrow2\left(x-y\right)^2+8\left(x-y\right)+8\le0\)
\(\Leftrightarrow2\left(x-y+2\right)^2\le0\)
Dấu = xảy ra khi \(\hept{\begin{cases}x+1=y-1\\x-y+2=0\end{cases}\Leftrightarrow}y=x+2\)
Thế vào P ta được
\(P=x^4+\left(x+2\right)^2-5x-5\left(x+2\right)+2020\)
\(=x^4+2x^2-6x+2014\)
\(=\left(x^2-1\right)^2+3\left(x-1\right)^2+2010\ge2010\)
Vậy GTNN là P = 2010 đạt được khi x = 1, y = 3
Ta có: √x+1+√y−1≤√2(x+y)
⇔√2(x−y)2+10x−6y+8≤√2(x+y)
⇔2(x−y)+10x−6y+8≤2(x+y)
⇔2(x−y)2+8(x−y)+8≤0
⇔2(x−y+2)2≤0
Dấu = xảy ra khi {
⇔y=x+2
Thế vào P ta được
P=x4+(x+2)2−5x−5(x+2)+2020
=x4+2x2−6x+2014
=(x2−1)2+3(x−1)2+2010≥2010
Vậy GTNN là P = 2010 đạt được khi x = 1, y = 3