đk: x;y;z dương nhé

áp dụng bđt cosi ta có:

\(x^2+yz>=2\sqrt{x^2yz}=2x\sqrt{yz};y^2+xz>=2\sqrt{y^2xz}=2y\sqrt{xz};z^2+xy=2\sqrt{z^2xy}=2z\sqrt{xy}\)

\(\Rightarrow\frac{1}{x^2+yz}< =\frac{1}{2x\sqrt{yz}};\frac{1}{y^2+xz}< =\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}< =\frac{1}{2z\sqrt{xy}}\)

\(\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(1\right)\)

áp dụng bđt cosi ta có:

\(\frac{1}{xy}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{xz}}=\frac{2}{x\sqrt{yz}};\frac{1}{xy}+\frac{1}{yz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{yz}}=\frac{2}{y\sqrt{xz}};\)

\(\frac{1}{yz}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{yz}\cdot\frac{1}{xz}}=\frac{2}{z\sqrt{xy}}\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{yz}+\frac{1}{xz}=\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}>=\frac{2}{x\sqrt{yz}}+\frac{2}{y\sqrt{xz}}+\frac{2}{z\sqrt{xy}}\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}>=\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)>=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(2\right)\)

từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}>=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\left(đpcm\right)\)

dấu = xảy ra khi x=y=z

nhầm từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)

\(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)

\(=y\left(\frac{1}{x}+\frac{1}{z}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}\)

\(=-1-1-1=-3\)

P+3=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)

P+3=\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0.\left(x+y+z\right)=0\)

=> P=\(-3\)

Chuc ban hoc tot

Ta có : \(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(\Rightarrow P+3=\frac{y+z}{x}+1+\frac{z+x}{y}+1+\frac{x+y}{z}+1\)

\(\Rightarrow P+3=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)

\(\Rightarrow P+3=\left(x+y+z\right).\frac{1}{x}+\left(x+y+z\right).\frac{1}{y}+\left(x+y+z\right).\frac{1}{z}\)

\(\Rightarrow P+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\Rightarrow P+3=\left(x+y+z\right).0\)

\(\Rightarrow P+3=0\)

\(\Rightarrow P=-3\)

Vậy P = - 3

*Xét trường hợp x+y+z = 0

Áp dụng tính chất dãy tỉ số bằng nhau
x/(y+z+1) = y/(x+z+1) = z/(x+y-2) = x+y+z/(y+z+1+x+z+1+x+y-2)=0

=>x=y=z=0

 *Xét x+y+z khác 0, tính chất tỉ lệ thức: 

x+y+z = x/(y+z+1) = y/(x+z+1) = z/(x+y-2) = (x+y+z)/(2x+2y+2z) = 1/2 
=> x+y+z = 1/2
+ 2x = y+z+1 = 1/2 - x + 1 => x = 1/2 
+ 2y = x+z+1 = 1/2 - y + 1 => y = 1/2 
+ z = 1/2 - (x+y) = 1/2 - 1 = -1/2 

Vậy có các cặp (x,y,z) thỏa mãn là: (0,0,0) và (1/2,1/2,-1/2)

b, \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

áp dụng dãy tỉ số bằng nhau :

\(\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

x = 2 . 10 = 20

y = 2 . 15 = 30

z = 2 . 21 = 42 

Vậy : ..... 

a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)

MSC của y là : 20

Có: \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

Áp dụng dãy tỉ số bằng nhau, ta có: 

\(2x+3y-z=186\)

\(\Rightarrow2.15+3.20-28=30+60-28=62\)

\(\frac{186}{62}=3\)

 x = 3 . 15 = 45

 y = 3 . 20 = 60

 z = 3 . 28 = 84

Vậy: ..... 

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{y+z+1}{x}=\frac{x+z+2019}{y}=\frac{x+y-2020}{z}=\frac{y+z+1+x+z+2019+x+y-2020}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow2=\frac{1}{x+y+z}\)\(\Rightarrow x+y+z=\frac{1}{2}\)

Ta có: 

​+) \(\frac{y+z+1}{x}=2\)\(\Rightarrow y+z+1=2x\)\(\Rightarrow x+y+z+1=3x\)\(\Rightarrow\frac{1}{2}+1=3x\)\(\Rightarrow3x=\frac{3}{2}\)\(\Rightarrow x=\frac{1}{2}\)

+) \(\frac{x+z+2019}{y}=2\)\(\Rightarrow x+z+2019=2y\)\(\Rightarrow x+y+z+2019=3y\)\(\Rightarrow\frac{1}{2}+2019=3y\)\(\Rightarrow3y=\frac{4039}{2}\)\(\Rightarrow y=\frac{4039}{6}\)

+) \(\frac{x+y-2020}{z}=2\)\(\Rightarrow x+y-2020=2z\)\(\Rightarrow x+y+z-2020=3z\)\(\Rightarrow\frac{1}{2}-2020=3z\)\(\Rightarrow3z=\frac{-4039}{2}\)\(\Rightarrow z=\frac{-4039}{6}\)

Lại có: \(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{4039}{6}\right)^{2017}+\left(\frac{-4039}{6}\right)^{2017}=4032+\left(\frac{4039}{6}\right)^{2017}-\left(\frac{4039}{6}\right)^{2017}=4032\)

+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)

+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)

\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)

+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)

Áp dụng bđt Bunhiacopxki

\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)

\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)

(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))

Tương tự: \(xy^2+yz^2+zx^2\le3\)

\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)

\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)

dvdfhfeye5

x : y : z : t = 2 : 3 : 4 : 5

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}=\frac{x+y+z+t}{2+3+4+5}=\frac{2}{7}\)

\(\Rightarrow x=\frac{2}{7}.2=\frac{4}{7};y=\frac{2}{7}.3=\frac{6}{7};z=\frac{2}{7}.4=\frac{8}{7};t=\frac{2}{7}.5=\frac{10}{7}\)

Ta có: \(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{49}{7}=7\)

\(\Rightarrow x=7.10=70;y=7.15=105;z=7.12=84\)

Dù nhầm nhưng cũng thank nha