Question

Cho x/2=y/3=z/5. Tìm x, y, z biết: x-2y+3z=38

Answer 1

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x-2y+3z}{2-2.3+3.5}=\frac{38}{11}$

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x=3811.2=7611y=3811.3=11411z=3811.5=19011

Answer 2

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

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

\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2-6+15}=\dfrac{38}{11}\)

\(\dfrac{x}{2}=\dfrac{38}{11}\Rightarrow x=\dfrac{76}{11}\)

\(\dfrac{y}{3}=\dfrac{38}{11}\Rightarrow y=\dfrac{114}{11}\)

\(\dfrac{z}{5}=\dfrac{38}{11}\Rightarrow z=\dfrac{190}{11}\)

Answer 3

Giải:

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2.3+3.5}=\dfrac{38}{11}\) 

Suy ra:

\(\dfrac{x}{2}=\dfrac{38}{11}\Rightarrow x=\dfrac{38.2}{11}=\dfrac{76}{11}\) 

\(\dfrac{y}{3}=\dfrac{38}{11}\Rightarrow y=\dfrac{38.3}{11}=\dfrac{114}{11}\) 

\(\dfrac{z}{5}=\dfrac{38}{11}\Rightarrow z=\dfrac{38.5}{11}=\dfrac{190}{11}\)

Answer 4

a) Áp dung TC của dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{-2y}{-6}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2-3+15}=\dfrac{x-2y+3z}{14}=\dfrac{38}{14}=\dfrac{19}{7}\)

Vậy\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{19}{7}\Leftrightarrow x=\dfrac{38}{7}\\\dfrac{y}{3}=\dfrac{19}{7}\Leftrightarrow y=\dfrac{57}{7}\\\dfrac{z}{5}=\dfrac{19}{7}\Leftrightarrow x=\dfrac{95}{7}\end{matrix}\right.\)

Kết luận:..

Answer 5

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

nên \(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{15}\)

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

\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2-6+15}=\dfrac{38}{19}=2\)

Do đó: x=4; y=6; z=10