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DT
0
HN
27 tháng 12 2018
\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\frac{1}{x^2}-\frac{2}{xy}+\frac{1}{y^2}\right]-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2}{xy}:\left[\frac{y^2-2.xy+x^2}{x^2y^2}\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}.\frac{x^2y^2}{x^2-2xy+y^2}-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2xy}{x^2-2xy+y^2}+\frac{-x^2-y^2}{x^2-2xy-y^2}\)
\(=\frac{2xy-x^2-y^2}{x^2-2xy+y^2}=\frac{-\left(x^2-2xy+y^2\right)}{x^2-2xy+y^2}=-1\)
28 tháng 12 2018
\(\frac{2011^3+11^3}{2011^3+2000^3}\)
\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)
\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)
\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)
\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)
đpcm
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2022
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)
\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)
Dấu "=" xảy ra khi \(x=y\)
\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)
\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)
\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)
\(B_{min}=1\) khi \(x=y\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 4 2020
\(C=\frac{\left(x+y+2\right)^2}{xy+2\left(x+y\right)}+\frac{xy+2\left(x+y\right)}{\left(x+y+2\right)^2}=\frac{8}{9}.\frac{\left(x+y+2\right)^2}{xy+2\left(x+y\right)}+\frac{\left(x+y+2\right)^2}{9\left(xy+2x+2y\right)}+\frac{xy+2x+2y}{\left(x+y+2\right)^2}\)
\(C\ge\frac{4}{9}.\frac{2x^2+2y^2+4xy+8x+8x+8}{xy+2x+2y}+2\sqrt{\frac{\left(x+y+2\right)^2\left(xy+2x+2y\right)}{9\left(xy+2x+2y\right)\left(x+y+2\right)^2}}\)
\(C\ge\frac{4}{9}.\frac{\left(x^2+y^2\right)+\left(x^2+4\right)+\left(y^2+4\right)+4xy+8x+8y}{xy+2x+2y}+\frac{2}{3}\)
\(C\ge\frac{4}{9}.\frac{2xy+4x+4y+4xy+8x+8y}{xy+2x+2y}+\frac{2}{3}\)
\(C\ge\frac{4}{9}.\frac{6\left(xy+2x+2y\right)}{xy+2x+2y}+\frac{2}{3}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)
\(C_{min}=\frac{10}{3}\) khi \(x=y=2\)
DT
13 tháng 6 2019
\(xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+2045.\)
\(=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+2045\)
\(=\left[\left(x^2-2x\right)\left(y^2+6y\right)+3\left(y^2+6y\right)\right]+12\left(x^2-2x+3\right)+2009.\)
\(=\left(x^2-2x+3\right)\left(y^2+6x\right)+12\left(x^2-2x+3\right)+2009\)
\(=\left(x^2-2x+3\right)\left(y^2+6x+12\right)+2009\)
\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\Leftrightarrow\left(x-1\right)^2+2\ge2\)
\(\left(y+3\right)^2\ge0\forall y\Leftrightarrow\left(y+3\right)^2+3\ge3\)
Suy ra \(B=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\ge2.3+2009=2015\)
Vậy GTNN của B=2015 khi x=1, y=-3.
21 tháng 7 2020
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
21 tháng 7 2020
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
16 tháng 9 2020
Sử dụng BĐT Cauchy Schwarz ta dễ có:
\(P=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)
\(\ge\frac{\left(x+y\right)^2}{x+y-2}\)
Ta cần chứng minh: \(\frac{\left(x+y\right)^2}{x+y-2}\ge8\)
\(\Leftrightarrow\left(x+y\right)^2-8\left(x+y\right)+16\ge0\)
\(\Leftrightarrow\left(x+y-4\right)^2\ge0\)( ĐPCM )
TT
16 tháng 9 2020
Có : \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)
Theo BĐT Cô - si ta có :
\(\frac{x^2}{y-1}+4\left(y-1\right)\ge2\sqrt{\frac{x^2}{y-1}.4\left(y-1\right)}=4x\)
\(\frac{y^2}{x-1}+4\left(x-1\right)\ge4y\)
Do đó ; \(\frac{x^2}{y-1}+\frac{y^2}{x-1}+4.\left(x+y-2\right)\ge4\left(x+y\right)\)
\(\Leftrightarrow\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge8\)
Hay : \(P\ge8\)
Dấu "=" xảy ra khi \(x=y=2\)
Vậy \(P_{min}=8\) khi \(x=y=2\)