Question

Tìm x,y,z biết \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)  và -x+y+z=-120

Answer 1

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

=>x=165,y=20,z=25

Answer 2

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

Answer 3

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

Answer 4

\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\) và \(-x+y+z=-120\)

Áp dụng t/c dãy tỉ số bằng nhau,ta có: \(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{-\frac{11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{-\frac{4}{3}}=90\)

Suy ra \(x=90.\frac{11}{6}=165\)

\(y=90.\frac{2}{9}=20\)

\(z=90.\frac{5}{18}=25\)

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