tìm x,y,z 5x=2y , 2x=3z và x.y=90

\(\frac{x}{2}=\frac{y}{5}=\frac{x}{3}=\frac{z}{2}\)và \(x.y=90\)

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

\(\Rightarrow\frac{x}{6}=\frac{y}{5}=\frac{z}{2}=\frac{x.y}{6.5}=\frac{90}{30}=3\)

\(\Rightarrow\frac{x}{6}=3\Rightarrow3.6=18\)

\(\frac{y}{5}=3\Rightarrow y=3.5=15\)

\(\frac{z}{2}=3\Rightarrow z=3.2=6\)

Vây x = 18 y = 15  z = 6

k nha ^-^

5x=2y => \(\frac{x}{2}=\frac{y}{5}=>\frac{x}{6}=\frac{y}{15}\)

2y=3z => \(\frac{y}{3}=\frac{z}{2}=>\frac{y}{15}=\frac{z}{10}\)

=> \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

=> \(\frac{x^2}{36}=\frac{x.y}{6.15}=\frac{90}{90}=1\)

=> x² =36

=> x= -6;6

Xet x=-6

=> y= 90: (-6)=-15

=> z= -15:15.10=-10

Xet x=6

=> y=90:6=15

=> z=15:15.10=10

Vậy ( x;y;z) =( -6;-15;-10) ; ( 6;15;10) 

Từ \(5x=2y\)\(\Rightarrow\frac{x}{y}=\frac{2}{5}\)

Từ \(2x=3z\)\(\Rightarrow\frac{x}{z}=\frac{3}{2}\)

Từ \(xy=90\)\(\Rightarrow x=\frac{90}{y};y=\frac{90}{x}\)

Ta có: \(\frac{x}{y}=\frac{2}{5}\)

Mà \(x=\frac{90}{y}\)

Nên \(\frac{\frac{90}{y}}{y}=\frac{2}{5}\)\(\Leftrightarrow\frac{90}{y^2}=\frac{2}{5}\)\(\Leftrightarrow y=\pm15\)

*Khi \(y=15\) thì \(x=\frac{90}{15}=6\) và \(z=\frac{6.2}{3}=4\)

*Khi \(y=-15\) thì \(x=\frac{90}{-15}=-6\) và \(z=\frac{-6.2}{3}=-4\)

Vậy \(\left\{x;y;z\right\}\in\left\{\left(6;15;4\right),\left(-6;-15;-4\right)\right\}\)

a)\(\left|x-2y\right|=5\Rightarrow\left[\begin{matrix}x-2y=5\\x-2y=-5\end{matrix}\right.\)

Từ \(2x=3y=5z\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)\(\Rightarrow\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}\)

Nếu x-2y=5

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}=\frac{x-2y}{15-20}=\frac{5}{-5}-1\)

\(\Rightarrow\left\{\begin{matrix}x=-15\\y=-10\\z=-6\end{matrix}\right.\)

Nếu x-2y=-5

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{2y}{20}=\frac{z}{6}=\frac{x-2y}{15-20}=\frac{-5}{-5}=1\)

\(\Rightarrow\left\{\begin{matrix}x=15\\y=10\\z=6\end{matrix}\right.\)

Vậy có 2 bộ (x,y,z). Đó là (-15;-10;-6), (15;10;6)

b) Từ \(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\)\(\Rightarrow\frac{x}{6}=\frac{y}{15}\left(1\right)\)

\(2x=3z\Rightarrow\frac{x}{3}=\frac{z}{2}\)\(\Rightarrow\frac{x}{6}=\frac{z}{4}\left(2\right)\)

Từ (1),(2)\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{4}\)

Đặt\(\)\(\frac{x}{6}=\frac{y}{15}=\frac{x}{4}=k\)

\(\Rightarrow\left\{\begin{matrix}x=6k\\y=15k\\z=4k\end{matrix}\right.\Rightarrow xy=90k^2\)

\(\Rightarrow90k^2=90\Rightarrow k^2=1\Rightarrow\left[\begin{matrix}k=1\\k=-1\end{matrix}\right.\)

Với k=1\(\Rightarrow\)\(\left\{\begin{matrix}x=6\\y=15\\z=4\end{matrix}\right.\)

Với k=-1\(\Rightarrow\left\{\begin{matrix}x=-6\\y=-15\\z=-4\end{matrix}\right.\)

Có :

\(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x}{6}=\frac{y}{15}\)

\(2x=3z\Rightarrow\frac{x}{3}=\frac{z}{2}\Rightarrow\frac{x}{6}=\frac{z}{4}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{4}\)

\(\Rightarrow x,y,z\)cùng dấu

Lại có : \(\Rightarrow\frac{x^2}{36}=\frac{y^2}{225}=\frac{z^2}{16}=\left(\frac{x}{6}\right)\left(\frac{y}{15}\right)=\frac{xy}{6.15}=\frac{90}{90}=1\)

\(\frac{x^2}{36}=1\Rightarrow x^2=36\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

\(\frac{y^2}{225}=1\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)

\(\frac{z^2}{16}=1\Rightarrow z^2=16\Rightarrow\orbr{\begin{cases}z=4\\z=-4\end{cases}}\)

Mà \(x,y,z\)cùng dấu

\(\Rightarrow\orbr{\begin{cases}x=6;y=15;z=4\\x=-6;y=-15;z=-4\end{cases}}\)

Vậy ...

Giải:
Ta có: 5x = 2y => x/2 = y/5 => x/6 = y/15

2x = 3z => x/3 = z/2 => x/6 = z/4

=> x/6 = y/15 = z/4

Đặt x/6 = y/15 = z/4 = k

=> x = 6k, y = 15k, z = 4k

Mà xy = 90

=> 6.k.15.k = 90

=> 90.k² = 90

=> k² = 1

=> k = 1 hoặc k = -1

+) k = 1 => x = 6, y = 15, z = 4

+) k = -1 => x = -6, y = -15, z = -4

Vậy x = 6, y = 15, z = 4 hoặc x = -6, y = -15, z = -4

câu trả lời rất dễ : do la mot so tu 0 den 100000000000000000000000000000000000000000000

a) Ta có:\(\frac{x}{4}=\frac{y}{5}\Rightarrow\frac{x^2}{16}=\frac{y^2}{25}=\frac{x.y}{20}=\frac{80}{20}=4\)

\(\Rightarrow\hept{\begin{cases}x^2=64\\y^2=100\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm8\\y=\pm10\end{cases}}\)

\(\frac{x}{4}=\frac{y}{5}\)nên x,y cùng dấu. Vậy\(\left(x;y\right)=\left(8;10\right);\left(-8;-10\right)\)

b)\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x}{15}=\frac{-3z}{6}=\frac{5x-y-3z}{15-5+6}=\frac{2}{16}=\frac{1}{8}\)

\(\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{5}{8}\\z=\frac{-2}{8}=\frac{-1}{4}\end{cases}}\)Vậy............................................

a) đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=4k\\y=5k\end{cases}}\)

=> x.y=4k.5k=20k²=80

20k²=80

k²=80:20

k²=4

=> k = 2

\(\hept{\begin{cases}x=4k=4.2=8\\y=5k=5.2=10\end{cases}}\)

vậy x=8 và y=10

b) Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x}{5.3}=\frac{y}{5}=\frac{3z}{3.\left(-2\right)}=\frac{5x-y-3z}{15-5-\left(-6\right)}=\frac{2}{16}=\frac{1}{8}\)

\(\frac{x}{3}=\frac{1}{8}\Rightarrow x=\frac{1}{8}.3=\frac{3}{8}\)

\(\frac{y}{5}=\frac{1}{8}\Rightarrow y=\frac{1}{8}.5=\frac{5}{8}\)

\(\frac{z}{-2}=\frac{1}{8}\Rightarrow z=\frac{1}{8}.\left(-2\right)=\frac{-1}{4}\)

Vậy ...

Theo đề, ta có

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\) và 5x-y+3z= 124

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

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\left(=\right)\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{124}{4}=31\)

=>  \(\frac{x}{3}=31\)

\(\frac{y}{5}=31\)

\(\frac{z}{-2}=31\)

=>  x = 93

y  = 155

z = -62

\(\frac{x}{3}\)\(=\)\(\frac{y}{5}\)\(=\)\(\frac{z}{-2}\) và  \(5x-y+3z=124\)

\(\frac{x}{3}\)\(=\)\(\frac{y}{5}\)\(=\)\(\frac{z}{-2}\)\(\left(=\right)\)\(\frac{5x}{15}\)\(=\)\(\frac{y}{5}\)\(=\)\(\frac{3z}{-6}\)\(=\)\(\frac{5x-y-3x}{15-5-\left(-6\right)}\)\(=\)\(\frac{124}{4}\)\(=\)\(31\)

\(\frac{x}{3}\)\(=\)\(31\)

\(\frac{y}{5}\)\(=\)\(31\)

\(\frac{x}{-2}\)\(=\)\(31\)

\(x=93\)

\(y=155\)

\(x=-62\)

\(F)\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\) và \(2x-y-z=49\)

Ta có: \(\frac{x}{2}=\frac{y}{4}\implies \frac{x}{10}=\frac{y}{15}\)

           \(\frac{y}{5}=\frac{z}{4}\implies\frac{y}{15}=\frac{z}{12} \)

Suy ra: \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{2x}{20}=\frac{2x-y-z}{20-15-12}=\frac{49}{-7}=-7\)

\(\implies \frac{x}{10}=-7\implies x=-70\)

         \(\frac{y}{15}=-7\implies y=-105\)

         \(\frac{z}{12}=-7\implies z=-84\)

Vậy \(x=-70;y=-105;z=-84\)

\(G) \frac{x}{2}=\frac{y}{4}\) và \(xy=2\)

Ta có: \(\frac{x}{2}=\frac{y}{4}\implies \frac{xy}{2}=\frac{y^2}{4}\)

\(\implies \frac{2}{2}=\frac{y^2}{4}\)

\(\implies y^2=2.4:2=4\)

\(\implies y=2=-2\)

\(+)y=2\implies x=1\)

\(+)y=-2\implies x=-1\)

Vậy có các cặp (x;y) là: \((1;2);(-1;-2)\)