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

\(\Rightarrow x=2k;y=3k;z=5k\)

\(\Rightarrow xyz=2k.3k.5k\)

\(\Rightarrow30k^3=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=\pm3\)

Khi k = 3 thì x = 3.2 = 6 ; y = 3.3 = 9; z = 3.5 = 15

Khi k = -3 thì x = (-3).2 = -6; y = (-3).3 = -9; z = (-3).5 = -15

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) và \(xyz=810\)

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

\(\Rightarrow x=2k;y=3k,z=5k\)

TA CÓ: xyz = 2k . 3k . 5k = 810

                  => ( 2.3.5 ) .k.k.k = 810

                  => 30 k³ = 810

                  => k³ = 810 : 30 = 27

=> k = 3

=> x=2k = 2.3 = 6

=>y=3k = 3.3=9

=> z=5k = 5.3 = 15

Vậy...

đặt k= x/2 =y/3=z/5
=> x=2k
    y=3k
    z=5k
=> xyz=2k3k5k=30k^3=810
=> k^3=810:30=27=3^3
=> k=3
=> x=6, y=9, z=15

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}vs\left(x.y.x=810\right)\)

Đặt : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=K\)

=> x=k.2

y=k.3

z=k.5

=> \(2k.3k.5k=810\)

\(\Rightarrow30k^3=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=\text{±3}\)

TH1 : k=3

=> x=3.2=6

y=3.3=9

x=3.5=15

TH2 : k=-3

=> x=-3.2=-6

y=-3.3=-9

x=-3.5=-15

Ta đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Rightarrow x=2k;y=3k;z=5k\)

\(\Rightarrow xyz=2k.3k.5k\)

\(\Rightarrow xyz=30.k^3\)

\(\Rightarrow810=30.k^3\)

\(\Rightarrow k^3=810:30=27\)

\(\Rightarrow k=3\)

Nên x = 3 x 2 = 6

       y = 3 x 3 = 9

       z = 3 x 5 = 15

Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow\left(\frac{x}{2}\right)^3=\left(\frac{y}{3}\right)^3=\left(\frac{z}{5}\right)^3=\frac{x}{2}.\frac{y}{3}.\frac{z}{5}=\frac{zyz}{2.3.5}=\frac{810}{30}=27\)

\(\Rightarrow\frac{x^3}{2^3}=\frac{y^3}{3^3}=\frac{z^3}{5^3}=\left(±3\right)^3\)

\(\Rightarrow\hept{\begin{cases}x^3=\left(±3\right)^3.2^3=\left(±6\right)^3\\y^3=\left(±3\right)^3.3^3=\left(±9\right)^3\\z^3=\left(±3\right)^3.5^3=\left(±15\right)^3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\pm6\\y=\pm9\\z=\pm15\end{cases}}\)

Mà x , y , z cùng dấu

=> ( x , y , z ) ∈ { ( -6 , -9 , -15 ) ; ( 6 , 9 , 15 ) }

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

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

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

cả nhà giúp mình nha

Đat 2x/3=3y/4=4z/5=k(k khác 0).                      Ta co x=3/2.k ; y=4/3k ; z =5/4k.                        => x+y+z=3/2k+4/3k+5/4k=k.(49/12)=49.     =>k=12 =>x=18;y=16 ;z=15

a.=>12x/18=12y/16=12z/15

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

12x/18=12y/16=12z/15=>12x+12y+12z/18+16+15=12*(x+y+z)/49=12*49/49

=>12x/18=12=>x=18

=>12y/16=12=>y=16

=>12z/15=15=>z=15

x=54, y=81, z=135

Cách làm như sau:

Nhân các tử vs nhau, các mẫu vs nhau ta đc xyz/2*3*5=810/30=27

=> x=27*2=...

     y=27*3=...

     z=27*5=...

Đặt x/2 = y/3 = z/5 là k .

=>x = 2k ;   y = 3k  ;   z = 5k.

Thay x=2k,y=3k,z=5k  vào x.y.z = 810 ta được :

     x.y.z = 810

hay 2k . 3k . 5k = 810

    30k³ = 810 

=> k³ = 27 

   k³ = 3³

=> k = 3 

=> x = 2.3 = 6 

     y = 3.3= 9

     z = 5.3 = 15

Theo tính chất dãy tỉ số bằng nhau.

Ta có x/2=y/3=z/5 và x+y+z=810

x/2=y/3=z/5=x+y+z=810/2*3*5=810/30=27

Do đó x/2=27 => x=27*2=54

          y/3=27 => y=27*3=81

          z/5=27 => z=27*5=135

ap dung tinh chat cua day ti so = nhau ta co 

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)\(=>\frac{x.y.z}{2.3.5}=\frac{810}{30}=27\)

\(=>\frac{x}{2}=27=>x=54\)

\(=>\frac{y}{3}=27=>y=81\)

\(=>\frac{z}{5}=27=>z=135\)

vay \(x=54\), \(y=81\), \(z=135\)

  x:2=y:3 => x=(2y)/3 (1) 
y:3= z:5 => y= (3z)/5(2) 
thế (2) vào (1) ra x=(6z)/15 
Có xyz=810 => ((6z)/15 x (3z)/5 x z)=810 => (6/25)z^3 -810=0 ( Bấm máy tính pt lập phương này ra) 
=> z=15, y=9, z=6 

Ta có:

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

Mà \(x\cdot y\cdot z=810\Leftrightarrow2k\cdot3k\cdot5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

Với \(k=3\Rightarrow x=6;y=9;z=15\)

Vậy x=6; y=9;z=15

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

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

\(\Rightarrow x.y.z=2k.3k.5k=30.k^3\)

\(\Rightarrow30.k^3=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

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

Vậy x=6, y=9 và z=15

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

\(\Rightarrow\)x=2.k ; y=3.k ; z=5.k

Suy ra:x.y.z=2k.3k.5k=810

\(\Rightarrow\)2k.3k.5k=810

\(\Rightarrow\)(2.3.5)\(^{_k3}\)=810

\(\Rightarrow k^3\)=810:30

\(\Rightarrow k^3\)=27:\(3^3\)

\(\Rightarrow\)k=3

Suy ra:x=3.2=6

y=3.3=9

z=3.5=15

Vậy x=6;y=9;z=15

Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) và \(x.y.z=810\)

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

\(\Rightarrow x=2k;y=3k;z=5k\)

Thay \(x=2k;y=3k;z=5k\) vào \(x.y.z=810\), ta được :

\(x.y.z=810\)

\(\Rightarrow\left(2k\right).\left(3k\right).\left(5k\right)=810\)

\(\Rightarrow30k^3=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k^3=3^3\)

\(\Rightarrow k=3\)

+ Nếu \(k=3\)

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

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