x=\(\dfrac{-4.\left(-10\right)}{8}=5\).

y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)

z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)

x = 5

y = 14

z = 12

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)

\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)

\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)

\(+)|t|-2=-54=>|t|=-52\)(vô lí)

\(+)|t|-2=54=>|t|=56=>t=\pm56\)

\(\dfrac{-4}{8}\) = \(\dfrac{x}{-10}\) ⇒ \(x\) =  - \(\dfrac{4}{8}\).(-10) = 5

y =  -7 : (\(-\dfrac{4}{8}\)) = 14

z = -\(\dfrac{4}{8}\) \(\times\) (-24) = 12

Vậy (\(x;y;z\)) = (5; 14; 12)

\(\left(x;y;z\right)=\left(5;14;12\right)\)

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

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

-x/3=24/4=6

=>x=-18

3/y²=6

=>y²=1/2

hay \(y=\pm\dfrac{\sqrt{2}}{2}\)

\(\dfrac{\left(z+3\right)^3}{-4}=6\)

=>(z+3)³=-24

\(\Leftrightarrow z+3=-\sqrt[3]{24}\)

hay \(z=-\sqrt[3]{24}-3\)

||t|-2|/8=6

=>||t|-2|=48

=>|t|-2=48

=>t=50 hoặc t=-50

Ta có: \(-\dfrac{12}{16}=-\dfrac{3}{4}\\ Vì:-\dfrac{12}{16}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-z}{8}\left(gt\right)\\ < =>-\dfrac{3}{4}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-z}{8}\\ =>x=\dfrac{15.4}{-3}=-20\\ y=\dfrac{-3.\left(-24\right)}{4}=18\\ z=-\dfrac{\left(-3\right).8}{4}=6\)

Ta có:\(\dfrac{-12}{16}\) =\(\dfrac{-3}{4}\)

mà \(\dfrac{-12}{16}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-Z}{8}\)

=>\(\dfrac{-3}{4}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-Z}{8}\)

=> -3.x=15.4=60

4.y=-3.(-24)=72

Z.(-4)=-3.8=-24

=>x=-20

y=18

Z=6

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