a,Ta có:\(2x+3y-2=186\Rightarrow2x+3y=188\)

AD t/c DTS bằng nhau ta có:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y}{2.15+3.20}=\frac{188}{90}=\frac{94}{45}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=\frac{94}{45}\Rightarrow x=\frac{94}{3}\\\frac{y}{20}=\frac{94}{45}\Rightarrow x=\frac{376}{9}\\\frac{z}{28}=\frac{94}{45}\Rightarrow x=\frac{2632}{45}\end{cases}}\)

b,Ta có:\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

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

AD t/c DTS bằng nhau ta có:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{18}=\frac{2x+3y-z}{2.15+3.20-18}=\frac{372}{62}=6\)

Tự tìm x

c,\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Tự áp dụng

cậu xem titan à

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\)

\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

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

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{y}{28}=\dfrac{2x+3y-z}{15\cdot2+3\cdot20-28}=\dfrac{186}{62}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=3\Rightarrow x=45\\\dfrac{y}{20}=3\Rightarrow y=60\\\dfrac{z}{28}=3\Rightarrow z=84\end{matrix}\right.\)

Vậy: ... 

$\genfrac{}{}{0ex}{}{x}{3}=\genfrac{}{}{0ex}{}{y}{4};\genfrac{}{}{0ex}{}{y}{5}=\genfrac{}{}{0ex}{}{z}{7}\Rightarrow \genfrac{}{}{0ex}{}{x}{15}=\genfrac{}{}{0ex}{}{y}{20};\genfrac{}{}{0ex}{}{y}{20}=\genfrac{}{}{0ex}{}{z}{28}$

Áp dụng tính chất dãy tỉ số bằng nhau :

$\genfrac{}{}{0ex}{}{x}{15}=\genfrac{}{}{0ex}{}{y}{20}=\genfrac{}{}{0ex}{}{z}{28}=\genfrac{}{}{0ex}{}{2x+3y-z}{30+60-28}=\genfrac{}{}{0ex}{}{186}{62}=3$

$\Rightarrow \genfrac{}{}{0ex}{}{x}{15}=3.15=45$

$\Rightarrow \genfrac{}{}{0ex}{}{y}{20}=3.20=60$

$\Rightarrow \genfrac{}{}{0ex}{}{z}{28}=3.28=84$

Vậy x = 45; y= 60; z = 84

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

Áp dụng tính chất dãy tỉ số bằng nhau :

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

\(\Rightarrow\frac{x}{15}=3.15=45\)

\(\Rightarrow\frac{y}{20}=3.20=60\)

\(\Rightarrow\frac{z}{28}=3.28=84\)

Vậy x = 45; y= 60; z = 84

Giải

Ta có:

x/3=y/4 và y/5=z/7

x/15=y/20 và y/20=z/28

x/15=y/20=z/28

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:x/15=y/20=z/28

Do x/15=y/20=z/28 nên suy ra: 2x/30=3y/60=z/28=2x+3y-z/30+60-28=186/62=3(Do 2x+3y-z=186)

Suy ra:

x/15=3 suy ra x=45

y/20=3 suy ra y=60

z/28=3 suy ra z=84

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

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

=>x/15=y/20=z/28=2x/30=3y/60

ap dung tinh chat day ti so bang nhau ta co:

2x+3y-z/30+60-28=186/62=3

=>2x/30=3=>2x=90=>x=45

=>3y/60=3=>3y=180=>y=60

=>z/28=3=>z=84

Ta có : x/y = 3/4 => x/3 = y/4 => x/15 = y/20    (1)

           y/z = 5/7 => y/5 = z/7 => y/20 = z/28     (2)

Từ (1) và (2) => x/15 = y/20 = z/28 = 2x/30 = 3y/60 = 2x + 3y - z/30 + 60 - 28 = 186/62 = 3

                     => x = 3.15 = 45

                          y = 3.20 = 60

                          z = 3.28 = 84

1)

Ta có:

\(2x=3y=4z\Leftrightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x-y-z}{\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}}=-420\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-420.\dfrac{1}{2}=-210\\y=-420.\dfrac{1}{3}=-140\\z=-420.\dfrac{1}{4}=-105\end{matrix}\right.\)

Vậy....

anh ơi còn câu 2 ạ

a, \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\&2x-3y+z=6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\right.\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\&2x-3y+z=6\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\\\dfrac{y}{12}=3\\\dfrac{z}{20}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)

Vậy, ...

b, \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\&2x+3y-z=186\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{28}\end{matrix}\right.\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\&2x+3y-z=186\)

Áp dụng t/c dãy tỉ số bằng nhau:

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=3\\\dfrac{y}{20}=3\\\dfrac{z}{28}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=45\\y=60\\z=84\end{matrix}\right.\)

Vậy, ...

c, Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)

\(\Rightarrow xyz=2k.3k.5k=1920\Rightarrow30k^3=1920\)

\(\Rightarrow k^3=64\Rightarrow k=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=3.4=12\\z=5.4=20\end{matrix}\right.\)

Vậy,...

a) x/3 = y/4 ; y/4 = z/5 và 2x - 3y + z = 6

<=> x/3 = y/4 <=> x/12 = y/16 (1)

<=> y/4 = z/5 <=> y/16 = z/20 (2)

Từ (1) và (2) suy ra : x/12 = y/16 = z/20

<=> 2x/24 = 3y/48 = z/20

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

2x/24 = 3y/48 = z/20 = 2x - 3y + z / 24 - 48 + 20 = -6/4 = -3/2

<=> x/3 = -3/2 => x = -9/2

<=> y/4 = -3/2 => y = -6

<=> z/5 = -3/2 => z = -15/2

Vậy x = -9/2 , b = -6 , z = -15/2 .

Answer:

1.

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

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

\(\Rightarrow2\frac{x}{30}=3\frac{y}{60}=\frac{z}{28}\)

Áp dụng tính chất của dãy tỷ số bằng nhau

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

\(\Rightarrow2\frac{x}{30}=3\Rightarrow x=45\)

\(\Rightarrow3\frac{y}{60}=3\Rightarrow y=60\)

\(\Rightarrow\frac{z}{28}=3\Rightarrow z=84\)

2.

Ta đặt: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)

\(\Rightarrow x=2k\)

\(\Rightarrow y=3k\)

\(\Rightarrow z=4k\)

\(\Rightarrow xyz=2k.3k.4k=24.k^3=648\)

\(\Rightarrow k^3=27\Rightarrow k=3\)

\(\Rightarrow\frac{x}{2}=3\Rightarrow x=6\)

\(\Rightarrow\frac{y}{3}=3\Rightarrow y=9\)

\(\Rightarrow\frac{z}{4}=3\Rightarrow z=12\)

3.

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

\(4x=2z\Rightarrow\frac{x}{2}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) và \(x+y+z=27\)

Áp dụng tính chất dãy tỉ số bằng nhau

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

\(\Rightarrow\frac{x}{2}=3\Rightarrow x=6\)

\(\Rightarrow\frac{y}{3}=3\Rightarrow y=9\)

\(\Rightarrow\frac{z}{4}=3\Rightarrow z=12\)