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

fnf tha

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

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

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

a) 3x = 2y \(\Rightarrow\)\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{2}.\frac{1}{5}=\frac{y}{3}.\frac{1}{5}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{3}=\frac{z}{7}.\frac{1}{3}\Rightarrow\frac{y}{15}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x+y+z}{10+15+21}=\frac{32}{46}=\frac{2}{3}\)

\(\hept{\begin{cases}x=10.\frac{2}{3}=\frac{20}{3}\\y=15.\frac{2}{3}=10\\z=21.\frac{2}{3}=14\end{cases}}\)

Vậy \(\hept{\begin{cases}x=10.\frac{2}{3}=\frac{20}{3}\\y=15.\frac{2}{3}=10\\z=21.\frac{2}{3}=14\end{cases}}\)

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

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

Ta có: 

\(xyz=810\\ \Rightarrow2k.3k.5k=810\\ \Rightarrow30k^3=810\\ \Rightarrow k^3=810:30\\ \Rightarrow k^3=27\\ \Rightarrow k=3\)

Vậy:

x = 2k = 2.3 = 6

y = 3k = 3.3 = 9

z = 5k = 5.3 = 15

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 ) }

Ta có:

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

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

=> \(\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{125}=\frac{810}{30}=27\)

=> \(\hept{\begin{cases}x^3=27.8=6^3\\y^3=27.27=9^3\\z^3=27.125=15^3\end{cases}}\)=> \(\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}\)

Vậy ...