làm hộ mik cho

a) Từ \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

Từ \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\) (2)

Từ (1) và (2) =>\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

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

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3\cdot9\\y=-3\cdot7\\z=-3\cdot3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

b) Từ \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) (1)

Từ \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2) =>\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

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

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

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

c) Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)

=> \(x=12k\) ; \(y=9k\) ;\(z=5k\)

=> xyz = \(12k\cdot9k\cdot5k\) =\(540\cdot k^3\) = 20

=>\(k^3=20:540=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\)

=>\(k=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\cdot12\\y=\dfrac{1}{3}\cdot9\\z=\dfrac{1}{3}\cdot5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=\dfrac{5}{3}\end{matrix}\right.\)

d) Từ \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)

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

\(\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{25+49+9}=\dfrac{585}{83}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{585}{83}\cdot25\\y^2=\dfrac{585}{83}\cdot49\\z^2=\dfrac{585}{83}\cdot9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=\\y^2=\\z^2=\end{matrix}\right.\) đề bài sai nên ko tìm được x ; y ; z

3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)

4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)

5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)

6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)

7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)

Câu 3:

\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)

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

\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)

Câu b:

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)

Câu c:

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

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)

Câu d:

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

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)

Câu e:

\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)

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

\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)

\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)