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a)Ta có: \(2x=3y;5y=7z\)và \(x-y-z=-27\)
\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)và\(x-y-z=-27\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)và \(x-y-z=-27\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y-z}{21-14-10}=\frac{-27}{-3}=9\)
Ta có:\(\frac{x}{21}=9\Rightarrow x=9.21=189\)
\(\frac{y}{14}=9\Rightarrow y=9.14=126\)
\(\frac{z}{10}=9\Rightarrow z=9.10=90\)
Vậy:\(x=189;y=126\)và\(z=90\)
b) \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)và\(x^2-2y^2+z^2=18\)
\(\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)và\(x^2-2y^2+z^2=18\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}=\frac{x^2-2y^2+z^2}{16-50+36}=\frac{18}{2}=9\)
Ta có:\(\frac{x^2}{16}=9\Rightarrow x^2=144\Rightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
\(\frac{2y^2}{50}=9\Rightarrow2y^2=450\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)
\(\frac{z^2}{36}=9\Rightarrow z^2=324\Rightarrow\orbr{\begin{cases}z=18\\z=-18\end{cases}}\)
Vậy: \(x=12;y=15;z=18\)hoặc \(x=-12;y=-15;z=-18\)
Ta có\(\frac{x}{-3}=\frac{y}{7}\Rightarrow\frac{x}{-3}.\frac{1}{-2}=\frac{y}{7}.\frac{1}{-2}\Rightarrow\frac{x}{6}=\frac{y}{-14}\left(1\right)\)
\(\frac{y}{-2}=\frac{z}{5}\Rightarrow\frac{y}{-2}.\frac{1}{7}=\frac{z}{5}.\frac{1}{7}\Rightarrow\frac{y}{-14}=\frac{z}{35}\left(2\right)\)
Từ (1)(2)
=> \(\frac{x}{6}=\frac{y}{-14}=\frac{z}{35}\)
=> \(\frac{-2x}{-12}=\frac{4y}{-56}=\frac{5z}{175}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{6}=\frac{y}{-14}=\frac{z}{35}=\frac{-2x}{-12}=\frac{4y}{-56}=\frac{5z}{175}=\frac{-2x-4y+5z}{-12+56+175}=\frac{146}{219}=\frac{2}{3}\)
=> \(\hept{\begin{cases}\frac{x}{6}=\frac{2}{3}\\\frac{y}{-14}=\frac{2}{3}\\\frac{z}{35}=\frac{2}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=-\frac{28}{3}\\z=\frac{70}{3}\end{cases}}\)
Bài làm:
Ta có: \(\frac{x}{-3}=\frac{y}{7}\Leftrightarrow\frac{x}{-6}=\frac{y}{14}\left(1\right)\)
và \(\frac{y}{-2}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{-35}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{-6}=\frac{y}{14}=\frac{z}{-35}\)
Áp dụng t/c của dãy tỉ số bằng nhau ta được:
\(\frac{x}{-6}=\frac{y}{14}=\frac{z}{-35}=\frac{-2x-4y+5z}{12-56-175}=\frac{146}{-219}=-\frac{2}{3}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{-6}=-\frac{2}{3}\\\frac{y}{14}=-\frac{2}{3}\\\frac{z}{-35}=-\frac{2}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=-\frac{28}{3}\\z=\frac{70}{3}\end{cases}}\)
Vậy \(x=4\) ; \(y=-\frac{28}{3}\) và \(z=\frac{70}{3}\)
1.
\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}=>\frac{y}{15}=\frac{z}{21}\)
=>\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
=> x=2x10=20
y=2x15=30
z=2x21=42
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
\(\frac{x}{-3}=\frac{y}{7}\)=> \(\frac{x}{-6}=\frac{y}{14}\)(1)
\(\frac{y}{-2}=\frac{z}{5}\)=> \(\frac{y}{14}=\frac{z}{-35}\)(2)
Từ (1), (2) => \(\frac{x}{-6}=\frac{y}{14}=\frac{z}{-35}\)và -2x - 4y + 5z = 146
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{-6}=\frac{y}{14}=\frac{z}{-35}=\frac{-2x-4y+5z}{-2.\left(-6\right)-4.14+5\left(-35\right)}=\frac{146}{-219}=-\frac{2}{3}\)
=> x = \(-\frac{2}{3}.\left(-6\right)\)= 4
y = \(-\frac{2}{3}.14\)= \(-\frac{28}{3}\)
z = \(-\frac{2}{3}.\left(-35\right)\)= \(\frac{70}{3}\)
=>x/6=y/-14
y/-14=z/35
=>x/6=y/-14=z/35
=>-2x/-12=4y/-56=5z/175
=>-2x-4y+5z/-12+56+175=146/219=2/3
=>x=4,y=-28/3,z=70/3