Question

Cho : \(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\)

Tìm x,y,z biết 2x-y+z = 27

Answer 1

\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{\left(6x-12y\right)+\left(8z-6x\right)+\left(12y-8z\right)}{4+9+16}=\dfrac{0}{29}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ =\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)

Answer 2

Answer 3

Ta có

2x−4y3=4z−3x2=3y−2z4

⇒3(2x−4y)3.3=2(4z−3x)2.2=4(3y−2z)4.4

⇒6x−12y32=8z−6x22=12y−8z42

=6x−12y+8z−6x+12y−8z32+22+42=0

Nên 2x−4y3=0⇒2x=4y⇒x4=y2(1)

Và4z−3x2=0⇒4z=3x⇒x4=z3(2)

Từ (1) và (2) suy ra x4=y2=z3⇒2x8=y2=z3=2x+z−y8+3−2=369=4

*x4=4⇒x=4.4=16

*y2=4⇒y=2.4=8

*z3=4⇒z=3.4=12

Vậy x = 16 và y = 8 và z = 12