\(b,\dfrac{x}{2}=\dfrac{y}{3}\) và \(x.y=54\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)
\(\Rightarrow x=9.2=18\)
\(y=9:3=3\)
a/ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Leftrightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12\left(x+y+z\right)}{12+18+15}=\dfrac{12.49}{49}=12\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\Leftrightarrow x=18\\\dfrac{12y}{16}=12\Leftrightarrow y=16\\\dfrac{12z}{15}=12\Leftrightarrow z=15\end{matrix}\right.\)
Vậy ..............
a, 2x/3 = 3y/4 => y = (4/3)(2x/3) = 8x/9
2x/3 = 4z/5 => z = (5/4)(2x/3) = 10x/12 = 5x/6
=> x + y + z = x + 8x/9 + 5x/6 = 49
hay là
(18 + 16 + 15)x/18 = 49, tức là x = 18
=> y = (8/9)18 = 16
và z = (5/6)18 = 15
b, Đặt x/2=y/3=k (k khác 0)
=> x=2k ; y=3k
x.y=54=>2k.3k=54=>6k2=54
suy ra k= \(\dfrac{+}{ }\)3
=>(x,y)=(6,9);(-6;-9)
