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x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
\(\dfrac{-4}{8}\) = \(\dfrac{x}{-10}\) ⇒ \(x\) = - \(\dfrac{4}{8}\).(-10) = 5
y = -7 : (\(-\dfrac{4}{8}\)) = 14
z = -\(\dfrac{4}{8}\) \(\times\) (-24) = 12
Vậy (\(x;y;z\)) = (5; 14; 12)
\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
a. \(\Rightarrow\left\{\begin{matrix}\dfrac{-10}{15}=\dfrac{x}{-9}\\\dfrac{-10}{15}=\dfrac{-8}{y}\\\dfrac{-10}{15}=\dfrac{z}{-21}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b. \(\Rightarrow\left\{\begin{matrix}\dfrac{-7}{6}=\dfrac{x}{18}\\\dfrac{-7}{6}=\dfrac{-98}{y}\\\dfrac{-7}{6}=\dfrac{-14}{z}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-21\\y=84\\z=-12\end{matrix}\right.\)
a) Ta có: \(\dfrac{-10}{15}=\dfrac{x}{-9}\)
\(\Rightarrow15x=-10.\left(-9\right)\)
\(\Rightarrow15x=90\)
\(\Rightarrow x=6\)
Khi đó: \(\dfrac{6}{-9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)
\(\Rightarrow y=\dfrac{-8\left(-9\right)}{6}=12\)
và \(z=\dfrac{-8\left(-21\right)}{12}\) \(=14\)
Vậy \(\left[{}\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b) Lại có: \(\dfrac{-7}{6}=\dfrac{x}{18}\)
\(\Rightarrow6x=-7.18\)
\(\Rightarrow6x=-126\)
\(\Rightarrow x=-21\)
Khi đó \(\dfrac{-21}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}\)
\(\Rightarrow y=\dfrac{-98.18}{-21}=84\)
và \(z=\dfrac{-14.84}{-98}=12\)
Vậy \(\left[{}\begin{matrix}x=-21\\y=84\\z=12\end{matrix}\right.\)
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
a ) \(7x=4y\) hay \(\dfrac{x}{4}=\dfrac{y}{7}\) và \(y-z=24\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{y-z}{7-4}=\dfrac{24}{3}=8\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=32\\y=56\end{matrix}\right.\)
Vậy ............
b ) \(\dfrac{x}{5}=\dfrac{y}{6},\dfrac{y}{8}=\dfrac{z}{7}\)
hay : \(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}\) và \(x+y-z=69\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Vậy .......
c.
\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)và x - y = 40
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x-y}{2-5}=\dfrac{40}{-3}\)
\(\dfrac{x}{2}=\dfrac{40}{-3}\Rightarrow x=\dfrac{40.2}{-3}=-\dfrac{80}{3}\)
\(\dfrac{y}{5}=\dfrac{40}{-3}\Rightarrow y=\dfrac{40.5}{-3}=-\dfrac{200}{3}\)
Vậy x = \(-\dfrac{80}{3}\), y = \(-\dfrac{200}{3}\)
Tương tự tiếp nghen 
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
Từ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\) và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\\\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\\\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\end{matrix}\right.\)
Theo bài ta có :
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)
\(x+y-z=10\)
\(\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8-12+15}=\dfrac{10}{5}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Leftrightarrow x=16\\\dfrac{y}{12}=2\Leftrightarrow y=24\\\dfrac{z}{15}=2\Leftrightarrow z=30\end{matrix}\right.\)
Vậy ....
\(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{24}\)
\(\Leftrightarrow\dfrac{-1}{2}=\dfrac{-7}{y}\)
\(\Rightarrow y=14\)
\(\Leftrightarrow\)\(\dfrac{1}{-2}=\dfrac{x}{-10}=\dfrac{z}{24}\)
\(\Rightarrow\) x=5
z=(-12)
Vậy: x=5; y=14; z=(-12)