a, Ta có : 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

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

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

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

mà 2x+3y-z=50

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

Phân thức số 2 có thật sự là $\frac{z}{y-2}$ không bạn? Bạn xem lại đề.

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-7=0

=>

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-4=0

=>k=1

=>x=4k-1=3; y=2k+2=4; z=3k-2=3-2=1

a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)

⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c: Ta có: 5x=8y=20z

nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)

Do đó: x=24; y=15; z=6

ok

okkkkkkkkkkkkkkk

a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

⇒2x = 3.30 = 90 ⇒ x = 45

3y = 3.60 = 180 ⇒ y = 60

z = 3.28 = 84

Ý b) có gì đó sai sai ?

c)Ta có :

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒x = 5.15 = 75

y = 5.10 = 50

z = 5.6 = 30

d)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)

⇒ x = 2k ; y = 3k ; z = 5k

⇒ xyz = 2k.3k.5k = 30k³ = 810

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