Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{6}=k\Leftrightarrow\hept{\begin{cases}x=3k\\y=4k\\z=6k\end{cases}}\)

Ta có \(3x^2+y^2-z^2=28\)

\(\Leftrightarrow3\left(3k\right)^2+\left(4k\right)^2-\left(6k\right)^2=28\)

\(\Leftrightarrow27k^2+16k^2-36k^2=28\)

\(\Leftrightarrow7k^2=28\)

\(\Leftrightarrow k^2=4\Rightarrow k=\pm2\)

Do đó 

 \(\frac{x}{3}=k\Leftrightarrow\frac{x}{3}=\pm2\Leftrightarrow\hept{\begin{cases}x=3.2=6\\x=3.\left(-2\right)=-6\end{cases}}\)

\(\frac{y}{4}=k\Leftrightarrow\frac{y}{4}=\pm2\Leftrightarrow\hept{\begin{cases}y=4.2=8\\y=4.\left(-2\right)=-8\end{cases}}\)

\(\frac{z}{6}=k\Leftrightarrow\frac{z}{6}=\pm2\Leftrightarrow\hept{\begin{cases}z=6.2=12\\z=6.\left(-2\right)=-12\end{cases}}\)

Vậy các cặp x,y,z thỏa mãn là \(\left\{x=6;y=8;z=12\right\}\left\{x=-6;y=-8;z=-12\right\}\)

ngu bỏ mẹ ra

bt rồi

gọi ba số đó lần lượt là x,y,z ta có

\(\hept{\begin{cases}x^3+y^3+z^3=-1009\\x:y=2:3\\x:z=4:9\end{cases}}\)thực hiện rút y và z ta có \(\hept{\begin{cases}y=\frac{3x}{2}\\z=\frac{9x}{4}\end{cases}}\)thế vào phunogw trình đầu tiên

\(x^3+\left(\frac{3x}{2}\right)^3+\left(\frac{9x}{4}\right)^3=-1009\Leftrightarrow x^3\left(1+\frac{27}{8}+\frac{729}{64}\right)=-1009\)

\(\Leftrightarrow\frac{x^3.1009}{64}=-1009\Leftrightarrow x^3=-64\Leftrightarrow x=-4\Rightarrow\hept{\begin{cases}y=-6\\z=-9\end{cases}}\)

Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2S-S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{100}}\)

\(\Rightarrow S=1-\frac{1}{2^{100}}\)

Vậy \(S=1-\frac{1}{2^{100}}\).

cảm ơn Lê Mai

\(\frac{8^2\cdot5^4}{2^5\cdot25}\)\(=\)\(\frac{2^6\cdot5^4}{2^5\cdot5^2}\)\(=2\cdot5^2=50\)

\(\frac{8^2.5^4}{2^5.25}=\frac{\left(2^3\right)^2.5^4}{2^5.5^2}=\frac{2^6.5^4}{2^5.5^2}=2.5^2=2.25=50\)

x² + x = 0

x (x+1 )= 0

=> x=0 ; x+1=0

                  x=-1

Vậy: S={0;-1}

2x-1² = 25

2x-1 = 25

2x = 25+1

2x = 26

x = 26:2

x = 13