cho \(\sqrt{2x+3}+\sqrt{y+3}=4\)
\(P=\sqrt{x+2}+\sqrt{y+9}\) đạt max tại \(x_1;y_1\) . tính :\(x_1^2+x_2^2\)
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XO
27 tháng 12 2021
a) ĐKXĐ : \(3\le x\le7\)
Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)
\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)
Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)
10 tháng 4 2023
\(\Leftrightarrow\left(\sqrt{x+2022}-\sqrt{y+2022}\right)+\left(x^3-y^3\right)=0\)
=>\(\dfrac{x-y}{\sqrt{x+2022}+\sqrt{y+2022}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)
=>x-y=0
=>x=y
P=2x^2-5x^2+x^2+12x+2023
=-2x^2+12x+2023
=-2(x^2-6x-2023/2)
=-2(x^2-6x+9-2041/2)
=-2(x-3)^2+2041<=2041
Dấu = xảy ra khi x=3
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 12 2018
ĐKXĐ: \(x\ge3;y\ge1\)
\(\sqrt{x-3}-\sqrt{y-1}+\sqrt[3]{x^2+x+1}-\sqrt[3]{y^2+5y+7}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{\sqrt{x-3}+\sqrt{y-1}}+\dfrac{x^2+x+1-y^2-5y-7}{\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}}=0\)
Để cho gọn gàng, ta đặt:
\(\left\{{}\begin{matrix}\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}=b>0\\\sqrt{x-3}+\sqrt{y-1}=a>0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-y^2-4y-4+x-y-2}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-\left(y+2\right)^2+\left(x-y-2\right)}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{\left(x-y-2\right)\left(x+y+3\right)}{b}=0\)
\(\Leftrightarrow\left(x-y-2\right)\left(\dfrac{1}{a}+\dfrac{x+y+3}{b}\right)=0\)
\(\Leftrightarrow x-y-2=0\) do \(\left\{{}\begin{matrix}x\ge3\\y\ge1\end{matrix}\right.\) \(\Rightarrow x+y+3>0\Rightarrow\dfrac{1}{a}+\dfrac{x+y+3}{b}>0\)
\(\Rightarrow x=y+2\)
Thay vào Q ta được:
\(Q=y^2-\left(y+2\right)^2+3\left(y+2\right)+4\sqrt{y}+4\)
\(\Rightarrow Q=-y+4\sqrt{y}+6=10-\left(y-4\sqrt{y}+4\right)=10-\left(\sqrt{y}-2\right)^2\le10\)
\(\Rightarrow Q_{max}=10\) khi \(\sqrt{y}-2=0\Rightarrow\left\{{}\begin{matrix}y=4\\x=6\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 10 2019
\(3-2P=\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{xz}}+\frac{z}{z+2\sqrt{xy}}\)
\(3-2P\ge\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)
\(\Rightarrow2P\le2\Rightarrow P\le1\)
Dấu "=" xảy ra khi \(x=y=z\)
\(M\le\sqrt{\left(1+1\right)\left(x+y+2\right)}=\sqrt{20}=4\sqrt{5}\)
\(M_{max}=4\sqrt{5}\) khi \(\left\{{}\begin{matrix}x-2=y+4\\x+y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
28 tháng 9 2021
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18-3x\\ \Leftrightarrow x^3-3x=18\\ y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\\ \Leftrightarrow y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{9-8}=6+3y\\ \Leftrightarrow y^3-3y=6\\ \Leftrightarrow P=x^3+y^3-3\left(x+y\right)+1993\\ P=x^3+y^3-3x-3y+1993=18+6+1993=2017\)
28 tháng 9 2021
Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}.x=18+3x\)
\(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)
\(\Rightarrow y^3=3-2\sqrt{2}+3+2\sqrt{2}+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}y=6+3y\)
\(P=x^3+y^3-3\left(x+y\right)+1993\)
\(=18+3x+6+3y-3x-3y+1993=2017\)