Question

tìm x,y,z biết
\(\frac{x}{3}=\frac{y}{4};\frac{z}{5}=\frac{y}{7}\)

và 2x+3y-z=106

Answer 1

Từ \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}.\frac{1}{7}=\frac{y}{4}.\frac{1}{7}\Leftrightarrow\frac{x}{21}=\frac{y}{28}\)

\(\frac{z}{5}=\frac{y}{7}\Rightarrow\frac{z}{20}=\frac{y}{24}\)

Theo t/c dãy tỉ số bằng nhau :

\(\Rightarrow\frac{x}{21}=\frac{y}{28}=\frac{z}{20}=\frac{2x+3y-z}{21.2+28.3-20}=\frac{106}{106}=1\)

\(\Rightarrow x=1.21=21;y=1.28=28;z=1.20=20\)

Answer 2

\(\frac{x}{3}=\frac{y}{4};\)\(\frac{z}{5}=\frac{y}{7}\)

suy ra:  \(\frac{x}{21}=\frac{y}{28}=\frac{z}{20}\)

hay   \(\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}\)

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

    \(\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}=\frac{2x+3y-z}{42+84-20}=1\)

suy ra:  \(\frac{2x}{42}=1\)\(\Rightarrow\)\(x=21\)

            \(\frac{3y}{84}=1\)  \(\Rightarrow\)\(y=28\)

            \(\frac{z}{20}=1\)\(\Rightarrow\)\(z=20\)

Answer 3

Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{21}=\frac{y}{28}\)

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

\(\Rightarrow\frac{x}{21}=\frac{y}{28}=\frac{z}{20}=\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}=\frac{2x+3y-z}{42+84-20}=\frac{106}{106}=1\)

\(\Rightarrow\hept{\begin{cases}x=21\\y=28\\z=20\end{cases}}\)

Answer 4

Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{21}=\frac{y}{28}\left(1\right)\)

           \(\frac{z}{5}=\frac{y}{7}\Rightarrow\frac{z}{20}=\frac{y}{28}\left(2\right)\)

Từ (1) và (2) ta có: \(\frac{x}{21}=\frac{y}{28}=\frac{z}{20}\) và \(2x+3y-z=106\)

Áp dụng t/c DTSBN ta có:

\(\frac{x}{21}=\frac{y}{28}=\frac{z}{20}=\frac{2x+3y-z}{2.21+2.28-20}=\frac{106}{78}=\frac{53}{39}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{21}=\frac{53}{39}\Rightarrow x=\frac{371}{13}\\\frac{y}{28}=\frac{53}{39}\Rightarrow y=\frac{1484}{39}\\\frac{z}{20}=\frac{53}{39}\Rightarrow z=\frac{1060}{39}\end{cases}}\)

Vậy ...

Answer 5

\(\text{Ta có : }\)

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{21}=\frac{y}{28}\)

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

\(\Rightarrow\frac{x}{21}=\frac{y}{28}=\frac{z}{20}=\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}=\frac{2x+3y-z}{42+84-20}=\frac{106}{106}=1\)

\(\Rightarrow\hept{\begin{cases}x=21\\y=28\\z=20\end{cases}}\)

\(\Rightarrow\text{♕}\Leftarrow\)

Answer 6

Ta có : \(\hept{\begin{cases}\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{21}=\frac{y}{28}\\\frac{z}{5}=\frac{y}{7}\Rightarrow\frac{z}{20}=\frac{y}{28}\end{cases}\Leftrightarrow\frac{x}{21}=\frac{y}{28}=\frac{z}{20}\left(1\right)}\)

Theo bài ra , ta có : \(\left(1\right)\Rightarrow\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}\)và \(2x+3y-z=106\)

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

\(\frac{2x}{42}=\frac{3y}{84}=\frac{z}{20}=\frac{2x+3y-z}{42+84-20}=\frac{106}{106}=1\)

\(\Rightarrow\hept{\begin{cases}x=21\\y=28\\z=20\end{cases}}\)