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Ta có : \(\frac{x}{y}=\frac{10}{9}\Rightarrow\frac{x}{10}=\frac{y}{9}\)(1)
\(\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{y}{3}=\frac{z}{4}\Leftrightarrow\frac{y}{9}=\frac{z}{12}\) (2)
Từ (1) và (2) => \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)
Ta có : \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)
Nên : \(\frac{x}{10}=6\Rightarrow x=60\)
\(\frac{y}{9}=6\Rightarrow y=54\)
\(\frac{z}{12}=6\Rightarrow z=72\)
Vậy x = 60 ; y = 54 ; z = 72
Ta có :\(\frac{x}{y}=\frac{10}{9};\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{3}=\frac{z}{4}\)
\(\Rightarrow\frac{x}{10}=\frac{3y}{27};\frac{9y}{27}=\frac{z}{4}\Rightarrow\frac{x}{30}=\frac{y}{27}=\frac{z}{36}\)và x-y+z=78
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x}{30}=\frac{y}{27}=\frac{z}{36}=\frac{x-y+z}{30-27+36}=\frac{78}{39}=2\)
Suy ra : \(\frac{x}{30}=2\Rightarrow x=60\)
\(\frac{y}{27}=2\Rightarrow y=54\)
\(\frac{z}{36}=2\Rightarrow z=72\)
Ta có : \(\frac{x}{y}=\frac{10}{9}\Rightarrow\frac{x}{10}=\frac{y}{9}\)
\(\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{y}{9}=\frac{z}{12}\)
Nên : \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)
Do đó : \(\frac{x}{10}=6\Rightarrow x=60\)
\(\frac{y}{9}=6\Rightarrow y=54\)
\(\frac{z}{12}=6\Rightarrow z=72\)
Vậy x = 60 ; y = 54 ; z = 72
\(\frac{x}{y}=\frac{10}{9}\text{ }\Rightarrow\text{ }\frac{x}{10}=\frac{y}{9}\text{ }\left(1\right)\)
\(\frac{y}{z}=\frac{3}{4}=\frac{9}{12}\text{ }\Rightarrow\text{ }\frac{y}{9}=\frac{z}{12}\text{ }\left(2\right)\)
Từ ( 1 ) và ( 2 ) suy ra : \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)
Do đó : x = 6 . 10 = 60 ; y = 6 . 9 = 54 ' z = 6 . 12 = 72
Ta có: \(\frac{x}{y}=\frac{10}{9}\) và \(\frac{y}{z}=\frac{3}{4}\)
=> \(\frac{y}{z}=\frac{9}{12}\)
=> \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)
=> \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{87}{13}=6\)
=>\(\frac{x}{10}=6=>x=60\) ; \(\frac{y}{9}=6=>y=54\); \(\frac{z}{12}=6=>z=72\)
vậy x=60 ; y=54 ; z=72
\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{9}{10}\\\dfrac{y}{z}=\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{y}{10}\\\dfrac{y}{4}=\dfrac{z}{3}\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{20}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{18}=\dfrac{y}{20}=\dfrac{z}{15}=\dfrac{x-y+z}{18-20+15}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.18=108\\y=6.20=120\\z=6.15=90\end{matrix}\right.\)
\(=>\dfrac{x}{10}=\dfrac{y}{9};\dfrac{y}{3}=\dfrac{z}{4}=>\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}\)
AD t/c của dãy tỉ số bằng nhau ta có
\(\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(=>\left[{}\begin{matrix}x=10.6=60\\y=9.6=36\\z=12.6=72\end{matrix}\right.\)
\(\frac{x}{y}=\frac{10}{9}\Rightarrow\frac{x}{10}=\frac{y}{9}\) (1)
\(\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{y}{3}.\frac{1}{3}=\frac{z}{4}.\frac{1}{3}\Rightarrow\frac{y}{9}=\frac{z}{12}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)và \(x-y+z=78\)Áp dụng TC DTSBN ta có :
\(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)
\(\Rightarrow x=60;y=54;z=72\)