Theo đề bài ta có :

X/3 =y/4 => x/15 = y/20

Y/5 = z/7 => y/20 = z/28

=> x/15 = y/20 = z/28

Và 2x/30 =3y/60 =z/28 biết 2x + 3y - z = 124

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

2x/30 = 3y/60 = z/28 = (2x+3y - z )/ 30 +60 - 28 = 124/62 = 2

* 2x/30 = 2 => 2x = 60 => x = 30

* 3y/60 = 2 => 3y = 120 => y = 40

* z/28 = 2 => z = 56

a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x

Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y

= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y

Ta lại có : 1 + 4y/24 = 1+4y / 9+3y

=> 24=9+3y => 15=3y => y=5

Vậy y=5

Nhớ like

b, 1+3y/12 = 1+5y/5x = 1+7y/4x

Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x

= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x

Ta lại có: 1+5y / 5x = 1+5y / 6+2x

=> 5x = 6+2x => 3x = 6 => x=2

Vậy x =2

Mình sửa lại câu a

1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x

= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x

Ta lại có: 1+4y/24 = 1+4y/ 9 +3x

=> 24 = 9 +3x => 15= 3x => x =5

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

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

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

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{3\left(x-1\right)}{6}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\)

\(\Leftrightarrow\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{3x-3+3y-6-z+3}{6+9-4}=\dfrac{\left(3x+3y-z\right)+\left(3-3-6\right)}{11}=\dfrac{50-6}{11}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=4\Leftrightarrow x=4.2+1=9\\\dfrac{y-2}{3}=4\Leftrightarrow y=4.3+2=14\\\dfrac{z-3}{4}=4\Leftrightarrow z=4.4+3=19\end{matrix}\right.\)

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:

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

d) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(xyz=810\)

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

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

Thay \(x=2k;y=3k;z=5k\) vào \(xyz=810\) ta được

\(2k.3k.5k=810\)

\(30k=810\)

\(k^3=27\)

=> k = 3

=> \(x=2.3=6\)

=> \(y=3.3=9\)

=> \(z=5.3=15\)

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

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

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

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

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

\(\Rightarrow\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)

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

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

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

Thay vào \(y+z+1=2x\) ; ta có :

\(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)

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

Thay vào \(x+z+2=2y\) ; ta có :

\(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

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

Thay vào \(x+y-3=2z\) ; ta có :

\(\dfrac{1}{2}-z-3=2z\Rightarrow3z=\dfrac{-5}{2}\Rightarrow z=\dfrac{-5}{6}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)

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