Và ???

đề bài là j vậy bn ??

\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

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

=>x/1=y/2=z/3=k

=>x=k; y=2k; z=3k

Ta có: x^2+y^2+z^2=14

=>k^2+4k^2+9k^2=14

=>14k^2=14

=>k^2=1

TH1: k=1

=>x=1; y=2; z=3

TH2: k=-1

=>x=-1; y=-2; z=-3

Ta có:

\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=4k\\z=6k\end{matrix}\right.\)

Mà x² + y² + z² = 14

=> (2k)² + (4k)² + (6k)² = 14

=> 4k² + 16k² + 36k² = 14

=> (4 + 16 + 36)k² = 14

=> 56k² = 14

\(\Rightarrow k^2=\dfrac{14}{56}=\dfrac{1}{4}\)

\(\Rightarrow k=\pm\dfrac{1}{2}\)

- Với \(k=\dfrac{1}{2}\) thì ta có:

\(x=2\cdot\dfrac{1}{2}=1\)

\(y=4\cdot\dfrac{1}{2}=2\)

\(z=6\cdot\dfrac{1}{2}=3\)

- Với \(k=-\dfrac{1}{2}\) thì ta có:

\(x=2\cdot\left(-\dfrac{1}{2}\right)=-1\)

\(y=4\cdot\left(-\dfrac{1}{2}\right)=-2\)

\(z=6\cdot\left(-\dfrac{1}{2}\right)=-3\)

Vậy x = 1, y = 2, z = 3 hoặc x = -1, y = -2, z = -3

Đề nghị bạn ghi rõ câu hỏi!

\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

\(\Leftrightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

Đặt : \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=4k\\z=6k\end{matrix}\right.\)

Mà \(2x^2+2y^2-z^2=1\)

\(\Leftrightarrow2.\left(2k\right)^2+2.\left(4k\right)^2-\left(6k\right)^2=1\)

\(\Leftrightarrow8k^2+32k^2-36k^2=1\)

\(\Leftrightarrow4k^2=1\)

\(\Leftrightarrow k^2=\dfrac{1}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\)

+) \(k=\dfrac{1}{2}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2.\dfrac{1}{2}=1\\y=4.\dfrac{1}{2}=2\\z=6.\dfrac{1}{2}=3\end{matrix}\right.\)

+) \(k=-\dfrac{1}{2}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}.2=-1\\y=-\dfrac{1}{2}.4=-2\\z=-\dfrac{1}{2}.6=-3\end{matrix}\right.\)

a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)

\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)

\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)

\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)

Xin lỗi mình chỉ làm được câu a)

1.Tìm x, y.

2.TÌM x, y, z.

3.TÌM x, y, z.

1) \(\dfrac{x}{3}=\dfrac{y}{4}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

\(\Rightarrow xy=12k^2=192\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm16\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\end{matrix}\right.\)

2) Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{-90}{9}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).2=-20\\y=\left(-10\right).3=-30\\z=\left(-10\right).5=-50\end{matrix}\right.\)

3) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.8=16\\z=2.5=10\end{matrix}\right.\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)

a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x

Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y

= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y

Ta lại có : 1 + 4y/24 = 1+4y / 9+3y

=> 24=9+3y => 15=3y => y=5

Vậy y=5

Nhớ like

b, 1+3y/12 = 1+5y/5x = 1+7y/4x

Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x

= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x

Ta lại có: 1+5y / 5x = 1+5y / 6+2x

=> 5x = 6+2x => 3x = 6 => x=2

Vậy x =2

Mình sửa lại câu a

1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x

= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x

Ta lại có: 1+4y/24 = 1+4y/ 9 +3x

=> 24 = 9 +3x => 15= 3x => x =5