Question

1, Tìm các số x,y,z, biết

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

b, \(\dfrac{2x}{3}\) = \(\dfrac{3y}{4}\) = \(\dfrac{4z}{5}\) và x + y + z = 49

Answer 1

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

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{20}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

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

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

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

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

\(=\dfrac{6}{2}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)

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

\(\Rightarrow x.\dfrac{2}{3}=y.\dfrac{3}{4}=z.\dfrac{4}{5}\)

\(\Rightarrow x:\dfrac{3}{2}=y:\dfrac{4}{3}=z:\dfrac{5}{4}\)

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

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

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

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

\(=\dfrac{49}{\dfrac{49}{12}}=12\)

\(\Rightarrow\left\{{}\begin{matrix}x=12.\dfrac{3}{2}=18\\y=12.\dfrac{4}{3}=16\\z=12.\dfrac{5}{4}=15\end{matrix}\right.\)

Answer 2

Ta có :

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

\(\dfrac{y}{3}=\dfrac{z}{5}=>\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)

Từ (1),(2)=>\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)=\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)

=>\(\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)

Answer 3

a)\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)

Áp dụng tc dãy tỉ ....

b)\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tc dãy tỉ ....

Answer 4

a, \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}\)

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

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}=\dfrac{2x-3y+z}{18-36+15}=\dfrac{6}{-3}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}2x=-2\cdot18=-36\\3y=-2\cdot36=-72\\z=-2\cdot15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-18\\y=-24\\z=-30\end{matrix}\right.\)

Vậy............................

b, \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng t/c của dts = nhau ta có:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

\(\Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)

Vậy.........

Answer 5

1, Tìm các số x,y,z, biết

a, \(Từ\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)=>\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{15}\)

=>\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{15}\)=\(\dfrac{2x-3y+z}{18-36+15}=\dfrac{6}{-3}=-2\)

=> x= -18;

=> y= -24;

=> z= -30

Vậy...

b, Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:\

\(\dfrac{x+y+z}{\dfrac{49}{12}}=\dfrac{49}{\dfrac{49}{12}}=12\)

=> x= 18

=> y= 16

=> z= 15

Vậy.....

GOOD LUCK!!!