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Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\)
nên \(\dfrac{x}{7}=\dfrac{y}{20}\)(1)
Ta có: \(\dfrac{y}{z}=\dfrac{5}{8}\)
nên \(\dfrac{y}{5}=\dfrac{z}{8}\)
hay \(\dfrac{y}{20}=\dfrac{z}{32}\)(2)
Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)
hay \(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)
mà 2x-5y+2z=100
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x-5y+2z}{14-100+64}=\dfrac{100}{-22}=\dfrac{-50}{11}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-50}{11}\\\dfrac{y}{20}=\dfrac{-50}{11}\\\dfrac{z}{32}=-\dfrac{50}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{350}{11}\\y=\dfrac{-1000}{11}\\z=\dfrac{-1600}{11}\end{matrix}\right.\)
Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\Rightarrow\dfrac{x}{14}=\dfrac{y}{40}\Rightarrow\dfrac{2x}{28}=\dfrac{5y}{200}\) \(\left(1\right)\)
Lại có: \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{40}=\dfrac{z}{64}\Rightarrow\dfrac{5y}{200}=\dfrac{2z}{128}\) \(\left(2\right)\)
Kết hợp ( 1 ) và ( 2 ) ta có: \(\dfrac{2x+5y-2z}{28+200-128}=\dfrac{100}{100}=1\)
⇒ \(\dfrac{2x}{28}=1\Rightarrow x=\dfrac{1.28}{2}=14\)
⇒ \(\dfrac{5y}{200}=1\Rightarrow y=\dfrac{1.200}{5}=40\)
⇒ \(\dfrac{2z}{128}=1\Rightarrow z=\dfrac{1.128}{2}=64\)
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\(\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\)
\(\frac{x}{35}=\frac{y}{100};\frac{y}{100}=\frac{z}{160}\Leftrightarrow\)\(\frac{x}{35}=\frac{y}{100}=\frac{z}{160}\Leftrightarrow\frac{2x+5y-2z}{70+500-320}=\frac{100}{250}=0,4\)
\(\Rightarrow x=0,4.35=14\)
\(y=0,4.100=40\)
\(z=0,4.160=64\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{y}=\frac{7}{20}\Leftrightarrow\frac{x}{7}=\frac{y}{20}\Leftrightarrow\frac{x}{14}=\frac{y}{40}\)
\(\frac{y}{z}=\frac{5}{8}\Leftrightarrow\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{y}{40}=\frac{z}{64}\)
\(\Leftrightarrow\frac{x}{14}=\frac{y}{40}=\frac{z}{64}=\frac{2x+5y-2z}{2.14+5.40-2.64}=\frac{100}{100}=1\)
\(\Leftrightarrow x=14\)
\(y=40\)
\(z=64\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{y}=\dfrac{7}{20}\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{20};\dfrac{z}{y}=\dfrac{5}{8}\Leftrightarrow\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{14}=\dfrac{y}{40}=\dfrac{z}{25}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{14}=\dfrac{y}{40}=\dfrac{z}{25}=\dfrac{2x+5y-2z}{14\cdot2+40\cdot5-2\cdot25}=\dfrac{100}{178}=\dfrac{50}{89}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{700}{89}\\y=\dfrac{2000}{89}\\z=\dfrac{1250}{89}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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ta có \(\frac{x}{y}=\frac{7}{20}\Rightarrow\frac{x}{7}=\frac{y}{20}\Rightarrow\frac{x}{14}=\frac{y}{40}\Rightarrow\frac{2x}{28}=\frac{5y}{200}\left(1\right)\)
\(\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{y}{40}=\frac{z}{64}\Rightarrow\frac{5y}{200}=\frac{2z}{128}\left(2\right)\)
\(\left(1\right)\&\left(2\right)\Rightarrow\frac{2x+5y-2z}{28+200-128}=\frac{100}{100}=1\)
\(\frac{2x}{28}=1\Rightarrow x=\frac{28.1}{2}=14\)
\(\frac{5y}{200}=1\Rightarrow y=\frac{200.1}{5}=40\)
\(\frac{2z}{128}=1\Rightarrow z=\frac{128.1}{2}=64\)
\(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{x}{35}=\frac{y}{100};\frac{y}{100}=\frac{z}{160}\Rightarrow\frac{x}{35}=\frac{y}{100}=\frac{z}{160}\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{35}=\frac{y}{100}=\frac{z}{160}=\frac{2x+5y-2z}{2.35+5.100-2.160}=\frac{100}{250}\)= số lẽ sai đề
\(\frac{x}{y}=\frac{7}{20}\Rightarrow x=\frac{7}{20}y\)
\(\frac{y}{z}=\frac{5}{8}\Rightarrow z=\frac{8}{5}y\)
\(2x+5y-2z=\frac{2.7}{20}y+5y-\frac{2.8}{5}y=\frac{5}{2}y=100\Leftrightarrow y=40\)
\(\Rightarrow x=\frac{7}{20}.40=14,z=\frac{8}{5}.40=64\).