Question

a,\(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{4}=\dfrac{z}{5}\)và x-y-z=28

b,\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\) và \(2x+3y-z=-14\)

c,\(3x=y,5y=4z\) và \(6x+7y+8z=456\)

d,\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)và x-y=4009

Cảm ơn trước nhé!❤❤❤⛇

Answer 1

a/

Theo đề,ta có:

+/ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

+/\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)\(\left(2\right)\)

Từ (1) và (2), ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

Do đó:

+/ \(\dfrac{x}{8}=\dfrac{28}{-19}\Rightarrow x=-\dfrac{224}{19}\)

+/\(\dfrac{y}{12}=\dfrac{28}{-19}\Rightarrow y=-\dfrac{336}{19}\)

+/\(\dfrac{z}{15}=\dfrac{28}{-19}\Rightarrow z=-\dfrac{420}{19}\)

Vậy: + \(x=-\dfrac{224}{19}\)

+ \(y=-\dfrac{336}{19}\)

+ \(z=-\dfrac{420}{19}\)

Answer 2

a,$\frac{x}{2}=\frac{y}{3},\frac{y}{4}=\frac{z}{5}$và x-y-z=28

Có \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

=>\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng tính chất DTSBN có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)=\(\dfrac{x-y-z}{8-12-15}=\dfrac{-28}{19}\)

=> x=\(\dfrac{-224}{19}\)

y=\(\dfrac{-336}{19}\)

z=\(\dfrac{-420}{19}\)

Answer 3

b/ Theo đề, ta có:

\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{z}{7}\) và \(2x+3y-z=-14\)

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

\(\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=-\dfrac{14}{14}=-1\)

Do đó:

+/ \(\dfrac{2x}{6}=-1\Rightarrow x=-3\)

+/ \(\dfrac{3y}{15}=-1\Rightarrow y=-5\)

+/ \(\dfrac{z}{7}=-1\Rightarrow z=-7\)

Vậy: + \(x=-3\)

+ \(y=-5\)

+ \(z=-7\)