Ta có: \(\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{2x}{14}=\frac{5y}{100}=\frac{2z}{64}=\frac{2x+5y-2z}{14+100-64}=\frac{100}{50}=2\)

\(\Rightarrow\frac{x}{7}=2\Rightarrow x=14;\frac{y}{20}=2\Rightarrow y=40;\frac{z}{32}=2\Rightarrow z=64\)

Tick nha Doan Hai My! 

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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\(\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\)

\(\frac{x}{35}=\frac{y}{100};\frac{y}{100}=\frac{z}{160}\Leftrightarrow\)\(\frac{x}{35}=\frac{y}{100}=\frac{z}{160}\Leftrightarrow\frac{2x+5y-2z}{70+500-320}=\frac{100}{250}=0,4\)

\(\Rightarrow x=0,4.35=14\)

\(y=0,4.100=40\)

\(z=0,4.160=64\)

\(\dfrac{x}{y}=\dfrac{7}{20}\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{20};\dfrac{z}{y}=\dfrac{5}{8}\Leftrightarrow\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{14}=\dfrac{y}{40}=\dfrac{z}{25}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{14}=\dfrac{y}{40}=\dfrac{z}{25}=\dfrac{2x+5y-2z}{14\cdot2+40\cdot5-2\cdot25}=\dfrac{100}{178}=\dfrac{50}{89}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{700}{89}\\y=\dfrac{2000}{89}\\z=\dfrac{1250}{89}\end{matrix}\right.\)

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

\(\frac{y}{z}=\frac{5}{8}\Rightarrow z=\frac{8}{5}y\)

\(2x+5y-2z=\frac{2.7}{20}y+5y-\frac{2.8}{5}y=\frac{5}{2}y=100\Leftrightarrow y=40\)

\(\Rightarrow x=\frac{7}{20}.40=14,z=\frac{8}{5}.40=64\).

Câu a) sai đề nhé bạn.

b) Ta có:

\(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\) và \(2x+5y-2z=100\)

\(\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}\) và \(2x+5y-2z=100\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{2x+5y-2z}{2.7+5.20-2.32}=\frac{100}{50}=2\)

\(\hept{\begin{cases}\frac{x}{7}=2\Rightarrow x=2.7=14\\\frac{y}{20}=2\Rightarrow y=2.20=40\\\frac{z}{32}=2\Rightarrow z=2.32=64\end{cases}}\)

Vậy \(x=14;y=40;z=64\)

k đi mk trả lời cho

\(\frac{x}{y}=\frac{7}{20}\Leftrightarrow\frac{x}{7}=\frac{y}{20}\Leftrightarrow\frac{x}{14}=\frac{y}{40}\)

\(\frac{y}{z}=\frac{5}{8}\Leftrightarrow\frac{y}{5}=\frac{z}{8}\Leftrightarrow\frac{y}{40}=\frac{z}{64}\)

\(\Leftrightarrow\frac{x}{14}=\frac{y}{40}=\frac{z}{64}=\frac{2x+5y-2z}{2.14+5.40-2.64}=\frac{100}{100}=1\)

\(\Leftrightarrow x=14\)

\(y=40\)

\(z=64\)

\(\frac{x}{y}=\frac{7}{20}=>\frac{x}{14}=\frac{y}{40}\)(1)

\(\frac{y}{z}=\frac{5}{8}=>\frac{y}{40}=\frac{z}{64}\)(2)

Từ (1) và (2) \(=>\frac{x}{14}=\frac{y}{40}=\frac{z}{64}\)

Áp dụng t/c của dãy tỉ số bằng nhau 

\(\frac{x}{14}=\frac{y}{40}=\frac{z}{64}=\frac{2x+5y-2z}{28+200-128}=\frac{100}{100}=1\)

\(=>\hept{\begin{cases}\frac{x}{14}=1=>x=14\\\frac{y}{40}=1=>y=40\\\frac{z}{64}=1=>z=64\end{cases}}\)

Vậy ...

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

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

\(\Rightarrow\frac{x}{7}=\frac{y}{20}=\frac{z}{32}\)

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

\(\frac{x}{7}=\frac{y}{20}=\frac{z}{32}=\frac{2x+5y-2z}{14+100-64}=\frac{100}{50}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{7}=2\\\frac{y}{20}=2\\\frac{z}{32}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\cdot7=14\\y=2\cdot20=40\\z=2\cdot32=64\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(14;40;64\right)\)