\(x-\sqrt{x+6}=\sqrt{y+6}-y\) x,y thuộc R
Tìm GTLN , GTNN của P = x+y
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\(x+y=\sqrt{x+6}+\sqrt{y+6}\ge0\Rightarrow x+y\ge0\)
\(x+y=\sqrt{x+6}+\sqrt{y+6}\le\sqrt{2\left(x+y+12\right)}\)
\(\Rightarrow\left(x+y\right)^2\le2\left(x+y+12\right)\)
\(\Rightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)
\(\Rightarrow x+y\le6\) (do \(x+y+4>0\))
\(P_{max}=6\) khi \(x=y=3\)
\(x+y=\sqrt{x+6}+\sqrt{y+6}\)
\(\Rightarrow\left(x+y\right)^2=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\ge x+y+12\)
\(\Rightarrow\left(x+y\right)^2-\left(x+y\right)-12\ge0\)
\(\Rightarrow\left(x+y+3\right)\left(x+y-4\right)\ge0\)
\(\Rightarrow x+y-4\ge0\) (do \(x+y+3>0\))
\(\Rightarrow x+y\ge4\)
\(P_{min}=4\) khi \(\left(x;y\right)=\left(-6;10\right)\) và hoán vị
Ta có: x - \(\sqrt{x+6}\) = \(\sqrt{y+6}\) - y (x; y \(\ge\) -6)
\(\Leftrightarrow\) P = x + y = \(\sqrt{x+6}+\sqrt{y+6}\)
\(\Leftrightarrow\) P2 = x + y + 12 + 2\(\sqrt{\left(x+6\right)\left(y+6\right)}\)
Áp dụng BĐT Cô-si cho 2 số ko âm x + 6 và y + 6 ta có:
\(x+y+12\ge2\sqrt{\left(x+6\right)\left(y+6\right)}\)
\(\Leftrightarrow\) P2 \(\le\) x + y + 12 + x + y + 12 = 2x + 2y + 24 = 2P + 24
\(\Leftrightarrow\) P2 - 2P - 24 \(\le\) 0
\(\Leftrightarrow\) P2 - 36 + 12 - 2P \(\le\) 0
\(\Leftrightarrow\) (P - 6)(P + 6) + 2(6 - P) \(\le\) 0
\(\Leftrightarrow\) (P - 6)(P + 4) \(\le\) 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}P-6\ge0\\P+4\le0\end{matrix}\right.\\\left\{{}\begin{matrix}P-6\le0\\P+4\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}-4\ge P\ge6\left(KTM\right)\\6\ge P\ge-4\left(TM\right)\end{matrix}\right.\)
\(\Rightarrow\) -4 \(\le\) P \(\le\) 6
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\(x-\sqrt{x+6}=\sqrt{y+6}-y\)
\(\Leftrightarrow x+y=\sqrt{x+6}+\sqrt{y+6}\)
Áp dụng BĐT Bu nhi a cốp xki ta có :
\(\left(x+y\right)^2=\left(\sqrt{x+6}+\sqrt{y+6}\right)^2\le2\left(x+y+12\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)-24\le0\)
\(\Leftrightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)
\(\Leftrightarrow-4\le x+y\le6\)
Vậy \(MIN_P=-4\) khi \(x=y=-2\) ; \(MAX_P=6\) khi \(x=y=3\)
+ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-3\\y\ge-4\end{matrix}\right.\)
\(gt\Rightarrow x+y=6\left(\sqrt{x+3}+\sqrt{4+y}\right)\le6\sqrt{2\left(x+y+7\right)}\)
\(\Rightarrow\left(x+y\right)^2\le72\left(x+y+7\right)\)
\(\Rightarrow\left(x+y\right)^2-72\left(x+y\right)-504\le0\)
\(\Rightarrow\left(x+y-36\right)^2\le1800\Rightarrow P\le36+30\sqrt{2}\)
max \(P=36+30\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+3=y+4\\x+y=36+30\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{37}{2}+15\sqrt{2}\\y=\frac{35}{2}+15\sqrt{2}\end{matrix}\right.\)
+ \(x+y=6\left(\sqrt{x+3}+\sqrt{y+4}\right)\)
\(\Rightarrow\left(x+y\right)^2=36\left(x+y+7+2\sqrt{\left(x+3\right)\left(y+4\right)}\right)\)
\(\Rightarrow\left(x+y\right)^2-36\left(x+y\right)-252=72\sqrt{\left(x+3\right)\left(y+4\right)}\ge0\)
\(\Rightarrow\left(x+y-42\right)\left(x+y+6\right)\ge0\Rightarrow x+y\ge42\)
Min \(P=42\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+3\right)\left(y+4\right)}=0\\x+y=42\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-3\\y=45\end{matrix}\right.\\\left\{{}\begin{matrix}x=46\\y=-4\end{matrix}\right.\end{matrix}\right.\)
1/ Điều kiện: x>=2009.
Ta có: \(y=x-2\sqrt{x-2009}=\left(x-2009\right)-2\sqrt{x-2009}+1+2008.\)
=> \(y=\left(\sqrt{x-2009}-1\right)^2+2008\)
Do \(\left(\sqrt{x-2009}-1\right)^2\ge0\) => \(y=\left(\sqrt{x-2009}-1\right)^2+2008\ge2008\)(Với mọi x>=2009)
GTNN của y là: y=2008
Đạt được khi \(\left(\sqrt{x-2009}-1\right)^2=0\) <=> x-2009=1 <=> x=2010
2/ Ta có: x+y=6 => y=6-x. Đặt A=x2y
=> A=x2y=x2(6-x)=6x2-x3 = x(6x-x2)=x(9-9+6x-x2)=x[9-(x2-6x+9)] =x[9-(x-3)2]
Do x>0 và (x-3)2 >=0 => A đạt giá trị lớn nhất khi (x-3)2=0 <=> x=3
=> GTLN của A=x2y là 3.9=27 Đạt được khi x=y=3
Bài 1: \(x+y+z+11=2\sqrt{x}+4\sqrt{y-1}+6\sqrt{z-2}\)
ĐKXĐ:\(x\ge0;y\ge1;z\ge2\)
\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\cdot\sqrt{y-1}\cdot2+4+\left(z-2\right)-2\cdot\sqrt{z-2}\cdot3+9=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2+\left(\sqrt{z-2}-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=2\\\sqrt{z-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=5\\z=11\end{matrix}\right.\)
Bài 2:
Q=|x+2|+|x-2|>=|x+2+2-x|=4
Dấu = xảy ra khi (x+2)(x-2)<=0
=>-2<=x<=2
ĐK: \(x\ge-6\);\(y\ge-6\)
Ta có: \(x-\sqrt{x+6}=\sqrt{y+6}-y\)
\(\Leftrightarrow x+y=\sqrt{x+6}+\sqrt{y+6}\)
\(\Leftrightarrow\left(x+y\right)^2=x+6+2\sqrt{\left(x+6\right)\left(y+6\right)}+y+6\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x+y\right)+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\)(*)
(*)\(2\sqrt{\left(x+6\right)\left(y+6\right)}\ge0\)
(*)\(\Leftrightarrow\left(x+y\right)^2=\left(x+y\right)+12\ge0\)
\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)-12\ge0\)
\(\Leftrightarrow\left(x+y-4\right)\left(x+y+3\right)\ge0\)
Mà \(x+y+3>0\)
\(\Rightarrow x+y-4>0\)
\(\Leftrightarrow x+y\ge4\)(1)
Áp dụng BĐT Cô si cho\(x+6\ge0;y\ge6\ge0\)
\(2\sqrt{\left(x+6\right)\left(y+6\right)}\le\left(x+6\right)\left(y+6\right)\)
\(\Leftrightarrow\left(x+y\right)^2=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\le x+y+12+x+6+y+6\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)-24\le0\)
\(\Leftrightarrow\left(x+y-6\right)\left(x+y+4\right)\le0\)
Mà \(x+y+4>0\)
\(\Rightarrow x+y-6\le0\)
\(\Leftrightarrow x+y\le6\)(2)
Từ (1) và (2)\(\Rightarrow4\le P\le6\)
Min P = 4\(\Leftrightarrow\left(x+6\right)\left(y+6\right)=0\Leftrightarrow\orbr{\begin{cases}x=-6\\y=-6\end{cases}}\)
\(x=-6\Rightarrow y=10\)
\(y=-6\Rightarrow x=10\)
Max P = 6\(\Leftrightarrow x=y=3\)
Vậy GTLN của P là 6 <=> x = y = 3
GTNN của P là 4 <=> x = -6 ; y = 10 hoặc x = 10 ; y = -6