Question

[chauchaubui]

Tìm x,y,x biết:

a, \(\dfrac{1+3y}{12}\)=\(\dfrac{1+5y}{5x}\)=\(\dfrac{1+7y}{4x}\)

b,ab=\(\dfrac{1}{2}\);bc=\(\dfrac{2}{3}\);ac=\(\dfrac{3}{4}\)

c,(x-1)=2(y-2),4(y-2)=3(z-3) và 2x+3y-z=50

Answer 1

[Nguyễn Lê Phước Thịnh]

b: \(ab\cdot bc\cdot ac=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow\left(abc\right)^2=\dfrac{1}{4}\)

Trường hợp 1: abc=1/2

\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{1}{2}:\dfrac{1}{2}=1\\a=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\\b=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)

Trường hợp 2: abc=-1/2

\(\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=-\dfrac{3}{4}\\b=-\dfrac{2}{3}\end{matrix}\right.\)

c: Theo đề, ta có: \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{1}\\\dfrac{y-2}{3}=\dfrac{z-3}{4}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{6}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot6-3\cdot6+3\cdot4}=\dfrac{45}{6}=\dfrac{15}{2}\)

Do đó: x-1=45; y-2=45/2; z-3=30

=>x=46; y=49/2; z=33