Question

x/2=y=z/3 và 3x+2y-4z=16

Answer 1

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3}=\dfrac{3x}{6}=\dfrac{2y}{2}=\dfrac{4z}{12}=\dfrac{3x+2y-4z}{6+2-12}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-8\\y=-4\\z=-12\end{matrix}\right.\)

Answer 2

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=y=\dfrac{z}{3}=\dfrac{3x}{6}=\dfrac{2y}{2}=\dfrac{4z}{12}=\dfrac{3x+2y-4z}{6+2-12}=\dfrac{16}{-4}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).2=-8\\y=\left(-4\right).1=-4\\z=\left(-4\right).3=-12\end{matrix}\right.\)

Answer 3

\(\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3}\)

⇒\(\dfrac{3x}{6}=\dfrac{2y}{2}=\dfrac{4z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x}{6}=\dfrac{2y}{2}=\dfrac{4z}{12}=\dfrac{3x+2y-4z}{6+2-12}=\dfrac{16}{-4}=-4\)

⇒\(\left\{{}\begin{matrix}x=-4.2=-8\\y=-4.1=-4\\z=-4.3=-12\end{matrix}\right.\)

Answer 4

x/2= y/1 = z/3.  A/d TCDTSBN :

x/2 = y/1 = z/3 = 3x+2y-4z/ 3.2+2-4.3 = 16/ -4 = -4

=> x = -8. y = -4. z = -12

 

Answer 5

bạn nhầm đề ở chỗ 3x+2y-4z=16 rồi nhé 

phải là 3x-2y+4z=16 nhé

Do x/2=z/3\(\Rightarrow\)x=2/3z

Ta có: 3x-2y+4z=16

\(\Rightarrow\)3.2/3z-2.z/3+4z=16

\(\Rightarrow\)2.z-2/3z+4z=16

\(\Rightarrow\)16/3.z=16

\(\Rightarrow\)z=16:16/3=3

\(\Rightarrow\)x=2/3:3=2

\(\Rightarrow\)y=3/3=1