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

ta có :

\(\frac{x}{3}=\frac{y}{5}\)

\(\frac{y}{4}=\frac{z}{3}\)

\(\Rightarrow\frac{x}{12}=\frac{y}{20}=\frac{z}{15}\)

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

\(\frac{x}{12}=\frac{y}{20}=\frac{z}{15}=\frac{4x}{48}=\frac{2z}{30}=\frac{4x-y+2z}{48-20+30}=\frac{116}{58}=2\)

\(\frac{x}{12}=3\Rightarrow x=36\)

\(\frac{y}{20}=2\Rightarrow y=40\)

\(\frac{z}{15}=2\Rightarrow z=30\)

a) \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\) (1)

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

Từ (1);(2) suy ra: \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

Theo đề: \(\left|x-2y\right|=5\)

\(\Rightarrow x-2y=5\) (nếu \(x-2y\ge0\Leftrightarrow x\ge2y\) )

    \(x-2y=-5\) (nếu \(x< 2y\) )

Vậy có hai trường hợp

TH1: Nếu \(x\ge2y\) suy ra: \(\frac{x}{15}=\frac{y}{10}\Rightarrow\frac{x}{15}=\frac{2y}{20}=\frac{x-2y}{15-20}=\frac{5}{-5}=-1\)

\(\Rightarrow\hept{\begin{cases}x=15.\left(-1\right)=-15\\y=10.\left(-1\right)=-10\\z=6.\left(-1\right)=-6\end{cases}}\) (nhận)

TH2: Nếu x < 2y suy ra: \(\frac{x}{15}=\frac{y}{10}\Rightarrow\frac{x}{15}=\frac{2y}{20}=\frac{x-2y}{15-20}=\frac{-5}{-5}=1\)

\(\Rightarrow\hept{\begin{cases}x=15.1=15\\y=10.1=10\\z=6.1=6\end{cases}}\) (nhận)

b) \(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\) (1)

    \(2x=3z\Rightarrow\frac{x}{3}=\frac{z}{2}\) (2)

Từ (1);(2) => \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k\)

\(\Rightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}\Rightarrow xy=6k.15k=90k^2=90\Rightarrow k^2=1\Rightarrow k=\left\{-1;1\right\}}\)

\(\Rightarrow\hept{\begin{cases}x=6.1=6\\y=15.1=15\\z=10.1=10\end{cases}}\) hoặc \(\hept{\begin{cases}x=6.\left(-1\right)=-6\\y=15.\left(-1\right)=-15\\z=10.\left(-1\right)=-10\end{cases}}\)

c) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

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

= \(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)

= \(\frac{2x+2y+2z}{x+y+z}\)

= \(\frac{2\left(x+y+z\right)}{x+y+z}=2\)

=> \(\frac{1}{x+y+z}=2\) => x + y + z = 1/2

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

                                       => y + z + x + 1 = 3x

                                       => 1/2 + 1 = 3x

                                      => 3/2 = 3x

                                      => x = 3/2 : 3 = 1/2

=> \(\frac{x+z+2}{y}=2\) => x + z + 2 = 2y

                                        => x + z + y + 2 = 3y

                                        => 1/2 + 2 = 3y

                                       => 5/2 = 3y

                                       => y = 5/2 : 3 = 5/6

=> \(\frac{x+y-3}{z}=2\)=> x + y - 3 = 2z

                                         => x + y + z - 3 = 3z

                                          => 1/2 - 3 = 3z

                                        => 3z = -5/2

                                         => z = -5/2 : 3 = -5/6

Vậy ...

a, P + 3x\(^{^2}\) - 4xy = 6y\(^{^2}\) - 9xy + x\(^2\)

=> P = 6y\(^2\)- 9xy + x\(^2\)+ 4xy - 3x\(^2\)= 6y\(^2\)- 5xy - 2x\(^2\)

=> P = 6y\(^2\) - 5xy - 2x\(^2\)

b, 

4y\(^2\) - 8xy - P = 5x\(^2\) - 12xy + 4y\(^2\)

=> P = 4y\(^2\) - 8xy - 5x\(^2\) + 12xy - 4y\(^2\) = 4xy - 5x\(^2\)

=> P = 4xy - 5x\(^2\)

c,

P - ( x\(^2\) - 2y\(^2\) + 3z\(^2\) ) + 3x\(^2\) - y\(^2\) + 2z\(^2\)= 2x\(^2\) - 3y\(^2\) -z\(^2\)

= P + 2x\(^2\) + y\(^2\) - z\(^2\) = 2x\(^2\) - 3y\(^2\) - z\(^2\)

=> P = 2x\(^2\) - 3y\(^2\) - z\(^2\) - 2x\(^2\) - y\(^2\) + z\(^2\)

=> P = -2y\(^2\)

#)Giải :

a) Ta có : \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)

\(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}\Rightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}}\)

Vậy x = 45; y = 60; z = 84

b) Áp dụng tính chất dãy tỉ số bằng nhau :

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

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

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(1\right)\\x+z+2=2y\left(2\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y-3=2z\left(3\right)\\x+y+z=\frac{1}{2}\left(4\right)\end{cases}}\)

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

Thay (1) vào (+) ta được :

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

\(\left(+_2\right)x+y+z=\frac{1}{2}\Rightarrow x+z=\frac{1}{2}-y\)

Thay (2) và (+₂) ta được :

\(\frac{1}{2}-y+2=2y\Rightarrow\frac{5}{2}=3y\Rightarrow y=\frac{5}{6}\)

\(\left(+_3\right)x+y+z=\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Rightarrow\frac{4}{3}+z=\frac{1}{2}\Rightarrow z=\frac{-5}{6}\)

Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}\)

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Rightarrow x=2k;y=3k;z=5k\)

\(\Rightarrow xyz=2k\cdot3k\cdot5k=30k^3\)

Mà \(xyz=810\Rightarrow30k^3=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

Thay vào tìm x,,z.

a) ta có: \(-3x=5y\Rightarrow\frac{x}{5}=\frac{y}{-3}\)

ADTCDTSBN

có: \(\frac{y}{-3}=\frac{x}{5}=\frac{y-x}{-3-5}=\frac{20}{-8}=\frac{5}{2}\)

=> y/-3 = 5/2 => y = -15/2

x/5 = 5/2 => x = 25/2

KL:...

b) ta có: \(\frac{2x}{3}=\frac{3y}{4}\Rightarrow8x=9y\Rightarrow\frac{x}{9}=\frac{y}{8}\)

\(\frac{3y}{4}=\frac{4z}{5}\Rightarrow15y=8z\Rightarrow\frac{y}{8}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{8}=\frac{z}{15}\)

ADTCDTSBN

có: \(\frac{x}{9}=\frac{y}{8}=\frac{z}{15}=\frac{x+y+z}{9+8+15}=\frac{49}{32}\)

=> x/9 = 49/32 => x = ...

...

Mãi mới nghĩ ra cách này:

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

Áp dụng BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

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

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

Thiết lập tương tự 2 BĐT còn lại và cộng theo vế,ta có:

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

\(=\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\) (đpcm)

Dẫu "=" xảy ra khi \(x=y=z\)

Dễ thôi bạn ơi\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}=\frac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}=\frac{x+y+z}{4\left(x+y+z\right)}=\frac{1}{4}\)

      Vì   \(\frac{1}{4}< \frac{3}{4}\)      

      \(\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{3}{4}\)

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