x^3/8 = y^3/64 = z^3/216 
=> (x/2)^3 = (y/4)^3 = (z/6)^3 
=> x/2 = y/4 = z/6 
=> x^2/4 = y^2/16 = z^2/36 = (x^2 + y^2 + z^2)/(4 + 16 + 36) = 14/56 = 1/4 (t.c dãy tỉ số bằng nhau) 
Suy ra : 
x^2 = 1 => x = 1 v x = -1 
y^2 = 4 => y = 2 v y = -2 
z^2 = 9 => z = 3 v z = -3

1) ADTCDTSBN, ta có:

 \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4

* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12

- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16

* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20

Vậy x = 12

       y = 16

       z = 20

x=12

y=16

z=20

\(A=x-2y+3z\left(x,y,z>0\right)\)

\(\left\{{}\begin{matrix}2x+4x+3z=8\left(1\right)\\3x+y-3z=2\left(2\right)\end{matrix}\right.\)

(1) <=> \(5x+5y=10\) <=> x+ y = 2

=> y = 2-x

Từ (1) => \(2x+4\left(2-x\right)+3z=8\) 

=> -2x +3z =0

=> \(x=\dfrac{3}{2}z\) => \(z=\dfrac{2}{3}x\) thay vào A

=> \(A=x-2\left(2-x\right)+3.\dfrac{2}{3}x=5x-4\ge-4\)

Vậy A_min = -4.

2x²+2y²+2z²=1 hay (2x)²+(2y)²+(2z)²=1

\(\text{Ta có:}\)\(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\)

\(\Rightarrow\frac{3}{8}x=\frac{3}{8}.\frac{y}{8}=\frac{3}{8}.\frac{z}{17}\)

\(\Rightarrow x=\frac{y}{8}=\frac{z}{27}\)

\(\text{Đặt:}\)\(x=\frac{y}{8}=\frac{z}{27}=k\)

\(\Rightarrow x=k\)

\(\frac{y}{8}=k\Rightarrow y=8k\)

\(\frac{z}{27}=k\Rightarrow z=27k\)

\(\text{Có:}\)\(2x^2-2y^2-z^2=1\)

\(\Rightarrow\left(2k\right)^2+2\left(8k^2\right)-27k^2=1\)

\(\Rightarrow k^2.\left(2+2.8^2-27^2\right)=1\)

\(\Rightarrow k^2.\left(-599\right)=1\)

\(\Rightarrow k^2=\frac{-1}{599}\left(\text{Vô lí}\right)\)

\(\Rightarrow x,y,z\text{ ko có gtrị}\)

khó + lười + nhiều = không làm

Hello

ko thích làm