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1/x + 1/y = 1/2018
<=> 1/x = 1/2018 - 1/y = (y - 2018)/(2018y)
<=> x = 2018y/(y - 2018)
=> x + y = 2018y/(y - 2018) + y = y^2/(y - 2018)
=> x - 2018 = 2018y/(y - 2018) - 2018 = 2018^2/(y - 2018)
=> P = 1
Đặt \(\sqrt{x^2+y^2}=c;\sqrt{y^2+z^2}=a;\sqrt{z^2+x^2}=b\)
Ta có:
\(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
\(=\frac{1}{2\sqrt{2}}\left(\frac{c^2+b^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{b^2+a^2-c^2}{c}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{\left(2a+2b+2c\right)^2}{2\left(a+b+c\right)}-2018\right)=\frac{1009}{\sqrt{2}}\)
đặt x-2016=a
y-2017=b
z-2018=c
ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)
=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)
=>\(a=b=c=4\)
còn lại tự lm nốt
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
Ta có \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2018}\Leftrightarrow\dfrac{x+y}{xy}=\dfrac{1}{2018}\Leftrightarrow2018x+2018y=xy\Leftrightarrow xy-2018x-2018y=0\Leftrightarrow xy-2018x-2018y+2018^2=2018^2\Leftrightarrow x\left(y-2018\right)-2018\left(y-2018\right)=2018^2\Leftrightarrow\left(x-2018\right)\left(y-2018\right)=2018^2\Leftrightarrow\sqrt{\left(x-2018\right)\left(y-2018\right)}=2018\Leftrightarrow2\sqrt{\left(x-2018\right)\left(y-2018\right)}=2.2018\Leftrightarrow x+y+2\sqrt{\left(x-2018\right)\left(y-2018\right)}=x+y+2.2018\Leftrightarrow x-2018+2\sqrt{\left(x-2018\right)\left(y-2018\right)}+y-2018=x+y\Leftrightarrow\left(\sqrt{x-2018}+\sqrt{y-2018}\right)^2=x+y\Leftrightarrow\sqrt{x-2018}+\sqrt{y-2018}=\sqrt{x+y}\Leftrightarrow\dfrac{\sqrt{x+y}}{\sqrt{x-2018}+\sqrt{y-2018}}=1\Leftrightarrow P=1\)
Vậy nếu \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2018}\) thì \(P=1\)
nhờ bạn gửi câu hỏi đúng lúc mà mình đỡ phải gửi
Ta có:
\(P=\frac{\sqrt{x+y}}{\sqrt{x-2018}+\sqrt{y-2018}}\)
\(\Leftrightarrow P^2=\frac{x+y}{x+y-4036+2\sqrt{\left(x-2018\right)\left(y-2018\right)}}\)
\(=\frac{x+y}{x+y-4036+2\sqrt{xy-2018x-2018y+2018^2}}\)
Mặt khác :
\(\frac{1}{x}+\frac{1}{y}=\frac{1}{2018}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{1}{2018}\)
\(\Leftrightarrow2018x+2018y=xy\)
\(\Leftrightarrow xy-2018x-2018y=0\)(1)
Thế (1) vào P^2 ta có :
\(P^2=\frac{x+y}{x+y-4036+2\sqrt{2018^2}}=\frac{x+y}{x+y}=1\)
\(\Rightarrow P=.......\)