Question

Tim x, y biêt:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}\) va \(x+y-z=54\)

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}\) va \(x+2y-3c=-20\)

\(5x=8y=20z\) va \(x-y-z=3\)

\(\dfrac{x}{3}=\dfrac{y}{4}\) va \(xy=48\)

Answer 1

1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

Answer 2

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

Answer 3

3,Từ \(5x=8y=20z\Rightarrow\dfrac{5x}{160}=\dfrac{8y}{160}=\dfrac{20z}{160}\)

\(\Rightarrow\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}\)

Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}=\dfrac{x-y-z}{32-20-8}=\dfrac{3}{4}\)

+\(\dfrac{x}{32}=\dfrac{3}{4}\Rightarrow x=\dfrac{32.3}{4}=24\)

+\(\dfrac{y}{20}=\dfrac{3}{4}\Rightarrow y=\dfrac{20.3}{4}=15\)

+\(\dfrac{z}{8}=\dfrac{3}{4}\Rightarrow z=\dfrac{3.8}{4}=6\)

Vậy \(x=24;y=15;z=6\)

Answer 4

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{6}=\frac{y}{5}=\frac{z}{3}=\frac{x+y-z}{6+5-3}=\frac{54}{8}=\frac{27}{4}\)

Answer 5

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\) (k≠0)

⇒x=3k

y=4k

Ta có:xy=48

⇒3k.4k=48

12.\(k^2=48\)

\(k^2=48:12=4\)

⇒ k∈{4;-4]

TH1:k=4

\(\Rightarrow\left[{}\begin{matrix}x=3k=3.4=12\\y=4k=4.4=16\end{matrix}\right.\)

TH2:k=-4

\(\Rightarrow\left[{}\begin{matrix}x=3k=3.\left(-4\right)=-12\\y=4k=4.\left(-4\right)=-16\end{matrix}\right.\)

Vậy x=12 thì y=16 ; x=-12 thì y=-17