Question

g*) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)  và x - 2y + 3z = 14

h*) \(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và x +y + z = 48

Answer 1

g) Đặt k = \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\)

=> \(\begin{cases}x-1=2k\\y-2=3k\\z-3=4k\end{cases}\)

=> \(\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}\)

=> x - 2y + 3z = 2k+1 - 6k - 4 + 12k + 9 = 8k + 6

=> 8k + 6 = 14

=> k = 1

=> \(\begin{cases}x=2\\y=5\\z=7\end{cases}\)

Answer 2

Nguyễn Huy Thắng Hoàng Lê Bảo Ngọc giúp câu h với 

Answer 3

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

\(\left[\begin{array}{nghiempt}12x-15y=0\\20z-12x=0\\15y-20z=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}12x=15y\\20z=12x\\15y=20z\end{array}\right.\)

\(\left[\begin{array}{nghiempt}\frac{x}{15}=\frac{y}{12}\\\frac{z}{12}=\frac{x}{20}\\\frac{y}{20}=\frac{z}{15}\end{array}\right.\)