Áp dụng tính chất dãy tỉ số bằng nhau :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x.y.z}{2.3.5}=\dfrac{-30}{30}=-1\\ =>\left\{{}\begin{matrix}x=\left(-1\right).2=-2\\y=\left(-1\right).3=-3\\z=\left(-1\right).5=-5\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x.y.z}{2.3.5}=\dfrac{-30}{30}=-1\)

\(+)\)\(\dfrac{x}{2}=-1\Rightarrow x=-1\times2=-2\)

\(+)\)\(\dfrac{y}{3}=-1\Rightarrow y=-1\times3=-3\)

\(+)\)\(\dfrac{z}{5}=-1\Rightarrow z=-1\times5=-5\)

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=

chết rùi nó bị lỗi

xin lỗi nha

ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

mà xyz = 30 => 2k.3k.5k = 30 => 30.k³ = 30 => k³ = 1 => k = 1

=> x = 2k => x = 2

y = 3k => y = 3

z = 5k => z = 5

KL:...

x : 2 = y : 3 = z : 5 và x+ y+ z = 30

x = 6 : 2 = 3 

y = 9 : 3 = 3 

z = 15 : 5 = 3 

6 : 2 = 9 : 3 = 15 : 5 

hok tốt

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Leftrightarrow x=2k;y=3k;z=5k\)

Có \(xyz=30\Leftrightarrow2k.3k.5k=30\)

\(\Rightarrow30k^3=30\)

\(\Rightarrow k=1\)

\(\Rightarrow x=1.2=2\)

\(y=1.3=3\)

\(z=1.5=5\)

a,7x=5y

=x/5=y/7

=x+y/5+7

=24/12

=2

b,x/2=y/3=z/5

=(x/2)³=(y/3)³=(z/5)³

=xyz/2.3.5

=-30/30

=-1

c,6x=4y=3z

=6x/12=4y/12=3z/12

=x/2=y/3=z/4

=x+y+z/2+3+4

=18/9

=2

k mik nha bn ^_^

Đặt  \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}=kak\left(kak\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}x=4kak\\y=3kak\\z=5kak\end{cases}}\)

Mà  \(x^2+y^2+z^2=200\)

\(\Leftrightarrow\left(4kak\right)^2+\left(3kak\right)^2+\left(5kak\right)^2=200\)

\(\Leftrightarrow16.kak^2+9.kak^2+25.kak^2=200\)

\(\Leftrightarrow kak^2.\left(16+9+25\right)=200\)

\(\Leftrightarrow kak^2.50=200\)

\(\Leftrightarrow kak^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}kak=2\\kak=-2\end{cases}}\)

+) Với  \(kak=2\)thì  \(\hept{\begin{cases}x=4kak=8\\y=3kak=6\\z=5kak=10\end{cases}}\)

+) Với  \(kak=-2\)thì  \(\hept{\begin{cases}x=4kak=-8\\y=3kak=-6\\z=5kak=-10\end{cases}}\)

Vậy ...

Đặt  \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Ta có :  \(xyz=-30\)

\(\Leftrightarrow2k\times3k\times5k=-30\)

\(\Leftrightarrow30k^3=-30\)

\(\Leftrightarrow k^3=-1\)

\(\Leftrightarrow k=-1\)

Thay vào ta được :

\(\hept{\begin{cases}x=2k=-2\\y=3k=-3\\z=5k=-5\end{cases}}\)

Vậy ...

\(b,\frac{x}{4}=\frac{y}{3}=\frac{z}{5}\)

\(\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{16+9+25}\)

\(=\frac{200}{50}=4\)

\(\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=4\)

Đến đây bn tính nốt nhé@_@

a: Đặt x/5=y/2=k

=>x=5k;y=2k

Ta có: xy=90

\(\Leftrightarrow10k^2=90\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

=>x=15; y=6

Trường hợp 2: k=-3

=>x=-15; y=-6

b: 4x=-5y

nên \(\dfrac{x}{-5}=\dfrac{y}{4}=k\)

=>x=-5k; y=4k

xy=-80

\(\Leftrightarrow-20k^2=-80\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

=>x=-10; y=8

Trường hợp 2: k=-2

=>x=10; y=-8

c: Đặt x/7=y/-2=k

=>x=7k; y=-2k

\(x^2y=-98\)

\(\Leftrightarrow49k^2\cdot\left(-2k\right)=-98\)

=>k=1

=>x=7; y=-2

d: Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=>x=2k; y=3k; z=5k

Ta có: xyz=-30

\(\Leftrightarrow30k^3=-30\)

=>k=-1

=>x=-2; y=-3; z=-5

xin lỗi mik ko bít câu này! Ngại quá!

a) \(2x=5y\)⇒\(x=\dfrac{5}{2}y\)⇒\(xy=\dfrac{5}{2}y^2\)

Thay \(xy=250\), ta có:

\(250=\dfrac{5}{2}y^2\)

⇒\(y^2=100\)⇒\(y=+-10\)

+) \(y=10\text{⇒}x=250:10=25\)

+) \(y=-10\text{⇒}x=250:-10=-25\)

\(a,2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=k\\ \Rightarrow x=5k;y=2k\\ xy=250\Rightarrow5k\cdot2k=250\Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=25;y=10\\x=-25;y=-10\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=a\Rightarrow x=3a;y=2a;z=4a\\ xyz=192\Rightarrow24a^3=192\Rightarrow a^3=8\Rightarrow a=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=4\\z=8\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=q\Rightarrow x=5q;y=2q;z=-3q\\ xyz=240\Rightarrow-30q^3=240\Rightarrow q^3=-8\Rightarrow q=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-10\\y=-4\\z=6\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}2x=5y\\xy=250\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5y=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2xy-5y^2=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y^2=500\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10\\x=25\end{matrix}\right.\)