Question

1, Tìm các số x, y, z biết 

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

b, \(\frac{y+z+1}{x}=\frac{z+x+4}{y}\) =\(\frac{x+y-5}{z}=\frac{1}{x+y+z}\)

Answer 1

a ) 

Ta có : 

\(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

\(\Rightarrow\frac{4\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)

\(\Rightarrow\frac{4+20y}{20x}=\frac{5+35y}{20x}\)

\(\Rightarrow4+20y=5+35y\)

\(\Rightarrow35y-20y=4-5\)

\(\Rightarrow15y=4-5\)

\(\Rightarrow15y=-1\)

\(\Rightarrow y=-\frac{1}{15}\)

Lại có : 

\(\frac{1+3y}{12}=\frac{1+5y}{5x}\)

\(\Rightarrow\frac{1+3.-\frac{1}{15}}{12}=\frac{1+5.-\frac{1}{15}}{5x}\)

\(\Rightarrow\frac{1-\frac{1}{5}}{12}=\frac{1-\frac{1}{3}}{5x}\)

\(\Rightarrow\frac{4}{5}:12=\frac{4}{3}:5x\)

\(\Rightarrow\frac{1}{15}=\frac{4}{3}:5x\)

\(\Rightarrow5x=\frac{4}{3}:\frac{1}{15}\)

\(\Rightarrow5x=20\)

\(\Rightarrow x=4\)

Vậy \(x=4;y=-\frac{1}{15}\)

Answer 2

a) Xét \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

\(\Rightarrow\frac{4x\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)

\(\Rightarrow4x\left(1+5y\right)=5\left(1+7y\right)\)

\(\Rightarrow4+20y=5+35y\)

\(\Rightarrow35y-20y=4-5\)

\(\Rightarrow15y=-1\)

\(\Rightarrow y=\frac{-1}{15}\)

Xét \(\frac{1+3y}{12}=\frac{1+5y}{5x}\)

\(\Rightarrow\frac{1+3.\frac{-1}{15}}{12}=\frac{1+5.\frac{-1}{15}}{5x}\)

\(\Rightarrow\frac{1+\frac{-1}{5}}{12}=\frac{1+\frac{-1}{3}}{5x}\)

\(\Rightarrow\frac{\frac{4}{5}}{12}=\frac{\frac{2}{3}}{5x}\)

\(\Rightarrow\frac{4}{5}:12=\frac{2}{3}:5x\)

\(\Rightarrow\frac{1}{15}=\frac{2}{3}:5x\)

\(\Rightarrow5x=\frac{2}{3}:\frac{1}{15}\)

\(\Rightarrow5x=\frac{30}{3}\)

\(\Rightarrow x=\frac{30}{3}:5\)

\(\Rightarrow x=\frac{30}{3}.\frac{1}{5}\)

\(\Rightarrow x=2\)

Vậy x = 2 ; y = \(\frac{-1}{15}\)

Answer 3

Hỏa Long Natsu sai phần tìm x rồi

x = 2 không phải bằng 4