a. Sai đề.

b. \(5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.5}{2}=\dfrac{15}{2}\\y=\dfrac{5.5}{2}=\dfrac{25}{2}\end{matrix}\right.\)

Vậy ...

a, Sai đề ko vậy

b, \(5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)

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

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

\(\Rightarrow x=\dfrac{5}{2}.3=\dfrac{15}{2}\)

\(y=\dfrac{5}{2}.5=\dfrac{25}{2}\)

a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).

Áp dụng t/c dãy tỉ số = nhau, ta có :

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

⇒2x = 3.30 = 90 ⇒ x = 45

3y = 3.60 = 180 ⇒ y = 60

z = 3.28 = 84

Ý b) có gì đó sai sai ?

c)Ta có :

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒x = 5.15 = 75

y = 5.10 = 50

z = 5.6 = 30

d)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)

⇒ x = 2k ; y = 3k ; z = 5k

⇒ xyz = 2k.3k.5k = 30k³ = 810

a)Ta có :

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

Mà x.y=3,6 => 2k+5k=3,6=>7k=3,6

Vậy k = \(\dfrac{18}{35}\)

\(x=2k\Rightarrow x=\dfrac{36}{35}\)

\(y=5k\Rightarrow y=\dfrac{18}{7}\)

\(a,\dfrac{x}{2}=\dfrac{y}{5}\)

\(\rightarrow\)\(x.5=y.2\)

\(x.x.5=y.x.2\)

\(x^2.5=3,6.2\)

\(x^2.5=7,2\)

\(x^2=1,44\)

\(\rightarrow x=1,2\) hoặc \(x=-1,2\)

Ý b bạn làm tường tự nha

\(\dfrac{x}{5}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}\)

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

\(\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=\dfrac{1}{4}\\\dfrac{y^2}{9}=\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\\\left[{}\begin{matrix}y=1,5\\y=-1,5\end{matrix}\right.\end{matrix}\right.\)

Vậy ..

a. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)

mà \(x.y=\dfrac{-5}{27}\)

hay \(-3k.5k=\dfrac{-5}{27}\)

\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)

\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)

Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)

Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)

Vậy.......

b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)

Vậy.........

1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

3,Từ \(5x=8y=20z\Rightarrow\dfrac{5x}{160}=\dfrac{8y}{160}=\dfrac{20z}{160}\)

\(\Rightarrow\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}\)

Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}=\dfrac{x-y-z}{32-20-8}=\dfrac{3}{4}\)

+\(\dfrac{x}{32}=\dfrac{3}{4}\Rightarrow x=\dfrac{32.3}{4}=24\)

+\(\dfrac{y}{20}=\dfrac{3}{4}\Rightarrow y=\dfrac{20.3}{4}=15\)

+\(\dfrac{z}{8}=\dfrac{3}{4}\Rightarrow z=\dfrac{3.8}{4}=6\)

Vậy \(x=24;y=15;z=6\)

Áp dụng tinshh chất dãy tỉ số bằng nhau ; ta được :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\)

Do đó :

\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)

\(\dfrac{y}{4}=2\Rightarrow y=2.4=8\)

\(\dfrac{z}{5}=2\Rightarrow z=2.5=10\)

Vậy x = 6 ; y = 8 ; z = 10

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

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

\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\) \

\(\Rightarrow x=2.3=6\)

\(y=2.4=8\)

\(z=2.5=10\)

Từ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)

Và \(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\)\(\dfrac{y}{12}=\dfrac{z}{16}\)

Suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=-1\Rightarrow x=-1\cdot9=-9\\\dfrac{y}{12}=-1\Rightarrow y=-1\cdot12=-12\\\dfrac{z}{16}=-1\Rightarrow z=-1\cdot16=-16\end{matrix}\right.\)

Ta có :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)(1)

\(\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{y}{12}=\dfrac{z}{16}\)(2)

Từ (1) và (2) , suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

Do đó :

\(\dfrac{x}{9}=-1\Rightarrow x=-1.9=-9\)

\(\dfrac{y}{12}=-1\Rightarrow y=-1.12=-12\)

\(\dfrac{z}{16}=-1\Rightarrow z=-1.16=-16\)

Vậy x = -9 ; y = -12 ; z = -16

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{54}=\dfrac{y}{24}=\dfrac{z}{32}\)

\(=\dfrac{3x-2y-z}{27-24-16}\)

\(=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=9.-1=-9\\y=12.-1=-12\\z=16.-1=-16\end{matrix}\right.\)

Ta có:

\(\dfrac{x}{19}=\dfrac{y}{21}\Leftrightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)

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

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{38}=2\Rightarrow x=38\\\dfrac{y}{21}=2\Rightarrow y=42\end{matrix}\right.\)

Vậy ..............

Chúc bạn học tốt!

\(\dfrac{x}{19}\)=\(\dfrac{y}{21}\) và 2x - y = 34

+ Ta có : \(\dfrac{x}{19}\)=\(\dfrac{y}{21}\)\(\Rightarrow\)\(\dfrac{2.x}{2.19}\)=\(\dfrac{y}{21}\)\(\Rightarrow\dfrac{2.x}{38}\)=\(\dfrac{y}{21}\) và 2x-y=34

+ Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{2.x}{38}\)=\(\dfrac{y}{21}\)=\(\dfrac{2.x-y}{38-21}\)=\(\dfrac{34}{17}\)=2

\(\Rightarrow\left\{{}\begin{matrix}x=2.38=76\\y=2.21=42\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)

Vậy x=38 và y=42 cần tìm.