d) \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)
=> \(\frac{y+z-x}{4+6-2}=\frac{8}{8}=1\)
=> \(\frac{x}{2}=1\Rightarrow x=2\)
=> \(\frac{y}{4}=1\Rightarrow y=4\)
=> \(\frac{z}{6}=1\Rightarrow z=6\)
b) \(\frac{x}{3}=\frac{y}{4}\Rightarrow x=y.\frac{3}{4}\)
\(\frac{y}{6}=\frac{z}{8}\Rightarrow z=y.\frac{8}{6}=y.\frac{4}{3}\)
=> \(3x-2y-z=y.3.\frac{3}{4}-2y-y.\frac{4}{3}=13\)
=> \(y.\frac{9}{4}-2y-y.\frac{4}{3}=y.\left(\frac{9}{4}-2-\frac{4}{3}\right)=13\)
=> \(y.\frac{-13}{12}=13\)
\(y=13:\frac{-13}{12}\)
\(y=-12\)
=> \(x=y.\frac{3}{4}=-9\)
=> \(z=y.\frac{4}{3}=-16\)
a) Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)=> x=3k ; y=4k ; z=5k
Ta có:
2x + 3y + 5z = 86
=> 2(3k) + 3(4k) + 5(5k) = 86
6k + 12k + 25k = 86
(6 + 12 + 25)k = 86
43k = 86
k = 86 : 43 = 2
Vậy x = 3k = 3 . 2 = 6
y = 4k = 4 . 2 = 8
z = 5k = 5 . 2 = 10
b) Ta có:
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)
\(\frac{y}{6}=\frac{z}{8}\Rightarrow\frac{y}{12}=\frac{z}{16}\)
Vậy \(\frac{x}{9}=\frac{y}{12}=\frac{z}{16}\)
Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{16}=k\)=> x=9k ; y=12k ; z=16k
Ta có:
3x - 2y - z = 13
=> 3(9k) - 2(12k) - 16k = 13
27k - 24k - 16k = 13
(27 - 24 - 16)k = 13
(-13)k = 13
k = 13 : (-13) = -1
Vậy x = 9k = 9 . (-1) = -9
y = 12k = 12 . (-1) = -12
z = 16k = 16 . (-1) = -16
c) Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)=> x=2k ; y=3k ; z=4k
Ta có: xy + yz + zx = 104
=> (2k)(3k) + (3k)(4k) + (4k)(2k) = 104
6k2 + 12k2 + 8k2 = 104
(6 + 12 + 8)k2 = 104
26k2 = 104
k2 = 104 : 26 = 4
=> k\(\in\){-2;2}
Vậy:
TH1: TH2:
x = 2k = 2 . (-2) = -4 x = 2k = 2 . 2 = 4
y = 3k = 3 . (-2) = -6 y = 3k = 3 . 2 = 6
z = 4k = 4 . (-2) = -8 z = 4k = 4 . 2 = 8
d) Ta có: \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\) và y+z-x=8
Theo tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{y+z-x}{4+6-2}=\frac{8}{8}=1\)
Vì \(\frac{x}{2}\)=1 => x=2.1=2
\(\frac{y}{4}\)=1 => y=4.1=4
\(\frac{z}{6}\)=1 => z=6.1=6
