Question

[nguyễn thị bích ngọc]

• x/2 bằng 2y/3 bằng 3z/4  và x.y.z bằng -108
• x+y+z bằng x/y+z-2 bằng y/z+x-3 bằng z/x+y+5
• \(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\)và 3x +5y+7z \(=\)123

Answer 1

[Lê Hà Phương]

a) \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\) và \(xyz=-108\)

Đặt: \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=k\)

\(\Rightarrow x=2k\)

 \(y=\frac{3}{2}k\)

\(z=\frac{4}{3}k\)

\(\Rightarrow xyz=2k.\frac{3}{2}k.\frac{4}{3}k=4k^3=-108\Rightarrow k^3=-27\Rightarrow k=\sqrt[3]{-27}=-3\)

Vậy:

\(x=2.\left(-3\right)=-6\) 

\(y=\frac{3}{2}.\left(-3\right)=-\frac{9}{2}\)

\(z=\frac{4}{3}.\left(-3\right)=-4\)

Answer 2

[Lê Hà Phương]

\(\frac{x}{y}=\frac{7}{20}\Leftrightarrow\frac{x}{7}=\frac{y}{20}\)

\(\frac{y}{z}=\frac{5}{8}\Leftrightarrow\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{y}{20}=\frac{z}{32}\)

\(\Rightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}\) và \(3x+5y+7z=123\)

ADTCCDTSBN, ta có:

\(\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{3x+5y+7z}{21+100+224}=\frac{123}{345}=\frac{41}{115}\)

\(\Rightarrow x=\frac{41}{115}.7=\frac{287}{115}\)

\(y=\frac{41}{115}.20=\frac{164}{23}\)

\(z=\frac{41}{115}.32=\frac{1312}{115}\)