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Có: \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=\sqrt{2019}\)
\(\Leftrightarrow\left[xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right]^2=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow\left[y\left(1+x^2\right)+x\left(1+y^2\right)\right]^2=2018\)
\(\Leftrightarrow y\left(1+x^2\right)+x\left(1+y^2\right)=\sqrt{2018}\)
hay \(A=\sqrt{2018}\)
\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)
áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương
ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)
ta có :
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)
lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :
\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)
\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)
vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673
Ta có:
\(VT=\sqrt{9x\left(xy-9x\right)}+\sqrt{9y\left(xy-9y\right)}\le\frac{9x+xy-9x}{2}+\frac{9y+xy-9y}{2}\)
\(=xy=VP\)
Dấu = xảy ra khi \(x=y=18\)
\(\Rightarrow S=\left(18-17\right)^{2018}+\left(18-19\right)^{2019}=1-1=0\)
Ta có:
VT=\sqrt{9x\left(xy-9x\right)}+\sqrt{9y\left(xy-9y\right)}\le\frac{9x+xy-9x}{2}+\frac{9y+xy-9y}{2}VT=9x(xy−9x)+9y(xy−9y)≤29x+xy−9x+29y+xy−9y
=xy=VP=xy=VP
Dấu = xảy ra khi x=y=18x=y=18
\Rightarrow S=\left(18-17\right)^{2018}+\left(18-19\right)^{2019}=1-1=0⇒S=(18−17)2018+(18−19)2019=1−1=0
\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+\sqrt{x^2+1}}=y+\sqrt{y^2+1}\\\frac{1}{y+\sqrt{y^2+1}}=x+\sqrt{x^2+1}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+\sqrt{x^2+1}=y+\sqrt{y^2+1}\left(1\right)\\-y+\sqrt{y^2+1}=x+\sqrt{x^2+1}\left(2\right)\end{cases}}\)
Cộng vế với vế của (1) và (2) ta có:
\(-2x-2y=0\Leftrightarrow-2\left(x+y\right)=0\Leftrightarrow x+y=0\)
\(\Rightarrow P=x^{2019}+y^{2019}=0\)
Lời giải:
$xy+\sqrt{(1+x^2)(1+y^2)}=1$
$\Leftrightarrow \sqrt{(1+x^2)(1+y^2)}=1-xy$
$\Rightarrow (1+x^2)(1+y^2)=(1-xy)^2$ (bp 2 vế)
$\Leftrightarrow x^2+y^2=-2xy$
$\Leftrightarrow (x+y)^2=0\Leftrightarrow x=-y$.
Khi đó:
$M=(x+\sqrt{1+(-x)^2})(-x+\sqrt{1+x^2})=(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x)$
$=1+x^2-x^2=1$
Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)
\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)
Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)
=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)
= xy+xz+xy+yz+xz+yz
=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)
=4038
Vậy P=4038
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.