1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

\(\dfrac{4x-3y}{5}=\dfrac{5y-4z}{3}=\dfrac{3z-5x}{4}\)

=>\(\left\{{}\begin{matrix}\dfrac{4x-3y}{5}=\dfrac{5y-4z}{3}\\\dfrac{4x-3y}{5}=\dfrac{3z-5x}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\left(4x-3y\right)=5\left(5y-4z\right)\\4\left(4x-3y\right)=5\left(3z-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-9y-25y+20z=0\\16x-12y-15z+25x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\end{matrix}\right.\)

mà x-y+z=200 nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}36x-102y+60z=0\\164x-48y-60z=0\\60x-60y+60z=12000\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}200x-150y=0\\-24x-42y=-12000\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-3y=0\\4x+7y=2000\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-10y=-2000\\4x-3y=0\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\4x=3y\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=200\\x=\dfrac{3}{4}y=150\\150-200+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\x=150\\z=250\end{matrix}\right.\)

4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)

Do đó: x=-16; y=-24; z=-30

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{-21}\)

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{-21}=\dfrac{x+y+z}{10+15+\left(-21\right)}=\dfrac{92}{14}=\dfrac{46}{7}\)

Còn lại bạn tự tính nha

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\)

⇒\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{2z}{-4}\)

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

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{2z}{-4}=\dfrac{x+y-2z}{5+1-\left(-4\right)}=\dfrac{160}{10}=16\)

⇒\(\left\{{}\begin{matrix}x=16.5=80\\y=16.1=16\\z=16.-2=-32\end{matrix}\right.\)

g,

\(\dfrac{3x-2y}{5}=\dfrac{2z-5x}{3}=\dfrac{5y-3z}{2}\)

\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}=\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}=0\)\(\Rightarrow3x-2y=2z-5x=5y-3z=0\)

* 3x - 2y = 0 \(\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)

* 2z - 5x = 0 \(\Rightarrow2z=5x\Rightarrow\dfrac{x}{2}=\dfrac{z}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{50}{10}=5\)

\(\cdot\dfrac{x}{2}=5\Rightarrow x=10\)

\(\cdot\dfrac{y}{3}=5\Rightarrow y=15\)

\(\cdot\dfrac{z}{5}=5\Rightarrow z=25\)

câu h thiếu điều kiện rồi bạn ơi

b) Ta có: 7x=10y=12z

nên \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}\)

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

\(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}=\dfrac{x+y+z}{\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{12}}=\dfrac{685}{\dfrac{137}{420}}=2100\)

Do đó:

\(\left\{{}\begin{matrix}x=2100\cdot\dfrac{1}{2}=1050\\y=2100\cdot\dfrac{1}{10}=210\\z=2100\cdot\dfrac{1}{12}=175\end{matrix}\right.\)

\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)

a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)

=> x=8

3y=18=>y=6

4z=72=>z=18

Vậy x=8 ; y=6 ; z=18

b, Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)

Câu c bạn làm tương tự nhé!

d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)

\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)

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

\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)

Vậy...

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)