Đáp án:

`d,`

Đặt  $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{6} = k$

`->` $x = 3k$ ; $y = 4k$ ; $z = 6k$ 

`->` $xyz = 3k . 4k . 6k = 576$

`->` $72k^3 = 576$ 

`->` $k^3 = 8 = 2^3$

`->` $k = 2$  

`=>` $\dfrac{x}{3} = k  → \dfrac{x}{3} = 2 → x = 6$

`=>` $\dfrac{y}{4} = k → \dfrac{x}{3} = 2 → y = 8$

`=>` $\dfrac{z}{6} = k → \dfrac{z}{6} = 2 → z = 12$

Vậy $(x ; y ; z) = ( 6 ; 8 ; 12)$

---------------

`e,`

Ta có:`3x = 5y = 10z`

`-> {3x}/30 = {5y}/30 = {10z}/30`

`-> x/10 = y/6 = z/3`

`-> x^2/100 = {2y^2}/{72} = z^2/9`

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

`x^2/100 = {2y^2}/{72} = z^2/9 = {x^2 - 2y^2 + z^2}/{100 - 72 + 9} = 148/37 = 4`

`=> x/10 = 4 -> x = 4 xx 10 = 40`

`=> y/6 = 4 -> y = 4 xx 6 = 24`

`=> z/3 = 4 -> z = 4 xx 3 = 12`

Vậy $(x ; y ; z) = ( 40 ; 24 ; 12)$

`∘` dariana

Đáp án:

Bạn tham khảo nhé.

Giải thích các bước giải:

$d)\,\dfrac x3=\dfrac y4=\dfrac z6=k\Rightarrow\left\{\begin{matrix}x=3k\\y=4k\\z=6k\end{matrix}\right.\\\Rightarrow xyz=3k.4k.6k=576\\\Rightarrow 72k^3=576\\\Rightarrow k^3=8\\\Rightarrow k^3=2^3\\\Rightarrow k=2\\\Rightarrow\left\{\begin{matrix}x=6\\y=8\\z=12\end{matrix}\right.$

Vậy $x=6,y=8,z=12$.

$e)\,3x=5y=10z\Rightarrow \dfrac{3x}{15}=\dfrac{5y}{15}=\dfrac{10z}{15}\\\Rightarrow \dfrac x5=\dfrac y3=\dfrac{z}{\dfrac32}=k\Rightarrow\left\{\begin{matrix}\dfrac x5=k\Rightarrow x=5k\\\dfrac y3=k\Rightarrow y=3k\\\dfrac z{\dfrac32}=k\Rightarrow z=\dfrac{3k}2\end{matrix}\right.\\\Rightarrow x^2-2y^2+z^2=\left(5k\right)^2-2.\left(3k\right)^2+\left(\dfrac{3k}2\right)^2=148\\\Rightarrow25k^2-2.9k^2+\dfrac{9k^2}4=148\\\Rightarrow k^2.\!\!\left(25-18+\dfrac94\right)=148\\\Rightarrow k^2\!\cdot\!\dfrac{37}4=148\\\Rightarrow k^2=16\\\Rightarrow k^2=\left(\pm4\right)^2\\\Rightarrow k=\pm4\\\Rightarrow\left\{\begin{matrix}x=\pm20\\y=\pm12\\z=\pm6\end{matrix}\right.$

Vậy $x=\pm20,y=\pm12,z=\pm6$.