Question

cho 5z - 6y/ 4 = 6x -4z /5 = 4y -5x / 6 và 3x -3y + 5z = 96  tìm x,y,z

Answer 1

\(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)

\(\Leftrightarrow\dfrac{4\left(5z-6y\right)}{16}=\dfrac{5\left(6x-4z\right)}{25}=\dfrac{6\left(4y-5x\right)}{36}\)

\(\Leftrightarrow\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}\)

ADTCDTSBN có:

\(\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}=\dfrac{20z-24y+30x-20z+24y-30x}{16+25+36}=0\)

Do đó \(20z-24y=0;30x-20z=0\)

\(\Leftrightarrow5z=6y;6x=4z\)

\(\Rightarrow y=\dfrac{5z}{6};x=\dfrac{4z}{6}\)

Có \(3x-3y+5z=96\Rightarrow3.\dfrac{4z}{6}-3.\dfrac{5z}{6}+5z=96\)

\(\Rightarrow z=\dfrac{64}{3}\) \(\Rightarrow y=\dfrac{160}{9}\)và \(x=\dfrac{128}{9}\)

Vậy...

Answer 2

cho x/3 = y/4 và y/5 = z/6. tìm M = 2x + 3y+ 4z / 3x + 4y + 5z