Ta có \(\frac{x+y+3z}{7}=\frac{y+z+3x}{8}=\frac{z+x+3y}{10}=\frac{x+y+3z+y+z+3x+z+x+3y}{7+8+10}\)

                                                                                              \(=\frac{5\left(x+y+z\right)}{25}=\frac{x+y+z}{5}=\frac{5}{x+y+z}\)(1)

Từ (1) => (x + y + z)² = 25 

=> \(\orbr{\begin{cases}x+y+z=5\\x+y+z=-5\end{cases}}\)

Khi x + y + z = 5 => \(\frac{5}{x+y+z}=1\)

=> \(\hept{\begin{cases}z+x+3y=10\\y+z+3x=8\\x+y+3z=7\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+2y=10\\x+y+z+2x=8\\x+y+z+2z=7\end{cases}}\Rightarrow\hept{\begin{cases}5+2y=10\\5+2x=8\\5+2z=7\end{cases}}\Rightarrow\hept{\begin{cases}y=2,5\\x=1,5\\z=1\end{cases}}\)(tm)

Khi x + y + z = -5 => \(\frac{5}{x+y+z}=-1\)

=> \(\hept{\begin{cases}x+y+3z=-7\\y+z+3x=-8\\z+x+3y=-10\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+2z=-7\\x+y+z+2x=-8\\x+y+z+2y=-10\end{cases}}\Rightarrow\hept{\begin{cases}-5+2z=-7\\-5+2x=-8\\-5+2y=-10\end{cases}}\Rightarrow\hept{\begin{cases}z=-1\\x=-1,5\\y=-2,5\end{cases}}\)(tm)

Vậy các cặp (x;y;z) thỏa mãn là (1,5;2,5;1) ; (-1,5;-2,5;-1) 

ta có 

​\(\frac{x}{3}\)=\(\frac{y}{2}\)=> \(\frac{x}{9}\)=\(\frac{y}{6}\)

\(\frac{y}{3}\)=\(\frac{z}{5}\)=>\(\frac{y}{6}\)=\(\frac{z}{10}\)

=>\(\frac{x}{9}\)=\(\frac{y}{6}\)=\(\frac{z}{10}\)

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

    \(\frac{x}{9}\)=\(\frac{y}{6}\)=\(\frac{z}{10}\)=> \(\frac{2x}{18}\)=\(\frac{y}{6}\)=\(\frac{3z}{30}\)=\(\frac{2x-y+3z}{18-6+30}\)=\(\frac{42}{42}\)=1

Ta lại có:

     \(\frac{2x}{18}\)= 1=> 2x=18=>x=9

       \(\frac{y}{6}\)= 1 =>y=6

      \(\frac{3z}{30}\)= 1=>3z=30=>z=10

 Vậy x=9 ; y=6 và z=10

Ta có bđt \(\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\)

\(\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)\)

Áp dụng nhiều lần bđt trên ta được

\(\(\frac{1}{3x+3y+2z}=\frac{1}{\left(2x+y+z\right)+\left(x+2y+z\right)}\le\frac{1}{4}\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}\right)\)\)

\(\(\le\frac{1}{4}\left(\frac{1}{\left(x+y\right)+\left(x+z\right)}+\frac{1}{\left(x+y\right)+\left(y+z\right)}\right)\)\)

\(\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\right]\)\)

\(\(\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)\)

C/m tương tự cho các bđt còn lại

\(\(\frac{1}{3x+2y+3z}\le\frac{1}{16}\left(\frac{2}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\)\)

\(\(\frac{1}{2x+3y+3z}\le\frac{1}{16}\left(\frac{2}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\right)\)\)

Cộng vế theo vế được

\(\(P\le\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)=\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{4}.6=\frac{3}{2}\)\)

Dấu "=" xảy ra

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{1}{2x}+\frac{1}{2x}+\frac{1}{2x=6}\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{3}{2x}=6\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x=\frac{1}{4}\end{cases}}\)\)

\(\(\Leftrightarrow x=y=z=\frac{1}{4}\)\)

Vậy ..........

cách khác :)) 

\(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\)\(\Leftrightarrow\)\(x+y+z\le3\)

\(P=\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\)

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

\(\ge\frac{9}{9\left(x+y+z\right)-\left(x+y+z\right)}=\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.3}=\frac{3}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)

0000....78

Ta có : \(\frac{x+y+z-3t}{t}=\frac{y+z+t-3x}{x}=\frac{z+t+x-3y}{y}=\frac{t+x+y-3z}{z}\)

=> \(\frac{x+y+z-3t}{t}+4=\frac{y+z+t-3x}{x}+4=\frac{x+z+t-3y}{y}+4=\frac{x+y+t-3z}{z}+4\)

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

=> \(\frac{2012}{x}=\frac{2012}{y}=\frac{2012}{z}=\frac{2012}{t}=\frac{2012+2012+2012+2012}{x+y+z+t}=\frac{2012.4}{2012}=4\)

=> x = y = z = t = 403

Khi đó A = x + 2y - 3z + t

              = x + 2x - 3x + x

             = x = 403

Vậy x = 403 

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

Có: \(\frac{y-2}{3}=\frac{2y-4}{6};\frac{z-3}{4}=\frac{3z-9}{12}\)

\(\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

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

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-2y+4+3z-9}{2-6+12}=\frac{14-6}{8}=\frac{8}{8}=1\)

Vì \(\frac{x-1}{2}=1\Rightarrow x-1=1.2=2\Rightarrow x=2+1=3\)

\(\frac{y-2}{3}=1\Rightarrow y-2=3.1=3\Rightarrow y=3+2=5\)

\(\frac{z-3}{4}=1\Rightarrow z-3=1.4=4\Rightarrow z=4+3=7\)

Tự kết luận

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)ta có :

\(\frac{16}{3x+3y+2z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)

\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)

\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)

Cộng theo vế 3 đẳng thức trên ta được :

\(16.\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)

\(\le4.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4.6=24\)

\(\Rightarrow\)\(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)

Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath

Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath