Cho x2 + y2 + 4/x2 =8
tìm min A =xy
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BH
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BH
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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
\(\dfrac{x^2+y^2}{2}\ge xy\Rightarrow-xy\ge-\dfrac{x^2+y^2}{2}\)
\(\Rightarrow4=x^2+y^2-xy\ge x^2+y^2-\dfrac{x^2+y^2}{2}=\dfrac{x^2+y^2}{2}\)
\(\Rightarrow x^2+y^2\le8\)
\(C_{max}=8\) khi \(x=y=\pm2\)
\(x^2+y^2\ge-2xy\Rightarrow-xy\le\dfrac{x^2+y^2}{2}\)
\(4=x^2+y^2-xy\le x^2+y^2+\dfrac{x^2+y^2}{2}=\dfrac{3}{2}\left(x^2+y^2\right)\)
\(\Rightarrow x^2+y^2\ge\dfrac{8}{3}\)
\(C_{min}=\dfrac{8}{3}\) khi \(\left(x;y\right)=\left(-\dfrac{2}{\sqrt{3}};\dfrac{2}{\sqrt{3}}\right);\left(\dfrac{2}{\sqrt{3}};-\dfrac{2}{\sqrt{3}}\right)\)
Đúng thì like giúp mik nha bạn. Thx bạn
A
1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
\(\dfrac{M}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{3\left(x^2+y^2+xy\right)-2\left(x^2+y^2+2xy\right)}{x^2+y^2+xy}=3-\dfrac{2\left(x+y\right)^2}{x^2+y^2+xy}\le3\)
\(\Rightarrow M\le9\)
\(M_{max}=9\) khi \(\left\{{}\begin{matrix}x+y=0\\x^2+y^2+xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-\sqrt{3};\sqrt{3}\right);\left(\sqrt{3};-\sqrt{3}\right)\)
\(\dfrac{M}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{\dfrac{1}{3}\left(x^2+y^2+xy\right)+\dfrac{2}{3}\left(x^2+y^2-2xy\right)}{x^2+y^2+xy}=\dfrac{1}{3}+\dfrac{2\left(x-y\right)^2}{3\left(x^2+y^2+xy\right)}\ge\dfrac{1}{3}\)
\(\Rightarrow M\ge1\)
\(M_{min}=1\) khi \(\left\{{}\begin{matrix}x-y=0\\x^2+y^2+xy=3\end{matrix}\right.\) \(\Rightarrow x=y=\pm1\)
LN
1
19 tháng 12 2020
A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
19 tháng 12 2020
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
H9
HT.Phong (9A5)
CTVHS
29 tháng 9 2023
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
ND
1
8 tháng 5 2021
Đặt \(P=\dfrac{xy}{xy+1}\Rightarrow\dfrac{1}{P}=\dfrac{xy+1}{xy}=1+\dfrac{1}{xy}\)
Ta có : \(xy\le\dfrac{x^2+y^2}{2}=\dfrac{8}{2}=4\Rightarrow\dfrac{1}{xy}\ge4\)
\(\Rightarrow\dfrac{1}{P}\ge5\Rightarrow P\le\dfrac{1}{5}\)
Dấu "=" xảy ra khi $x=y=2$
2 tháng 8 2021
1)
Ta có: x+y=2
nên \(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy=2\)
hay xy=1
Ta có: \(x^3+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=2^3-3\cdot1\cdot2\)
=2
2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)
\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)
HH
3
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 4 2022
\(\left(x^2+9\right)+\left(y^2+9\right)+3\left(x^2+y^2\right)\ge6x+6y+6xy=90\)
\(\Rightarrow4\left(x^2+y^2\right)+18\ge90\)
\(\Rightarrow x^2+y^2\ge18\)
\(P_{min}=18\) khi \(x=y=3\)
\(x+y+xy=15\Rightarrow\left\{{}\begin{matrix}x\le15\\y\le15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\left(x-15\right)\le0\\y\left(y-15\right)\le0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2\le15x+15y\) (1)
Cũng từ đó ta có: \(\left(x-15\right)\left(y-15\right)\ge0\Rightarrow xy\ge15x+15y-225\)
\(\Rightarrow16x+16y-225\le x+y+xy=15\)
\(\Rightarrow x+y\le15\) (2)
(1);(2) \(\Rightarrow x^2+y^2\le15.15=225\)
\(P_{max}=225\) khi \(\left(x;y\right)=\left(0;15\right);\left(15;0\right)\)
x2+y2+4/x2=8
=>x4+x2y2+4-8x2=0
=>x4-8x2+16=12-x2y2
=>(x2-4)2=12-x2y2
=>x2y2 ≤ 12 => |xy| ≤ \(\sqrt{12}=2\sqrt{3}\)
=>min xy \(\ge-2\sqrt{3}\)
xy min khi: x=2, y=\(-\sqrt{3}\)