Question

[Death]

Tìm x,y,z biết:

a) \(\frac{x-1}{2005}=\frac{3-y}{2006}\)và x-y=4009

b) 3x=y ; 5y=4z và 6x+7y+8z=456

giúp mình với                                                    

Answer 1

[Phía sau một cô gái]

a) \(\frac{x-1}{2005}=\frac{3-y}{2006}\)

\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{\left(x-1\right)+\left(3-y\right)}{2005+2006}=\frac{x-1+3-y}{4011}=\frac{4009-1+3}{4011}=\frac{4011}{4001}=1\left(tc\right)\)

\(\Rightarrow\frac{x-1}{2005}=1\rightarrow x-1=2005\leftrightarrow x+1=2006\)

\(\Rightarrow\frac{3-y}{2006}=1\rightarrow3-y=2006\leftrightarrow y=3-2006=-2003\)

=>Vậy:\(x=2006;y=-2003\)

b) \(3x=y\)

\(\rightarrow\frac{x}{1}=\frac{y}{3}\)

\(\frac{\rightarrow x}{4}=\frac{y}{12}\)

\(5y=4z\)

\(\rightarrow=\frac{y}{4}=\frac{z}{5}\)

\(\rightarrow\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\frac{6x}{24}=\frac{7y}{84}=\frac{8z}{120}\)

\(\frac{6x}{24}=\frac{7y}{84}=\frac{8z}{120}=\frac{6x+7y+8z}{24+84+120}=\frac{456}{228}=2\left(tc\right)\)

\(\cdot\frac{x}{4}=2\rightarrow x=4\cdot2=8\)

\(\cdot\frac{y}{12}=2\rightarrow y=12\cdot2=24\)

\(\cdot\frac{z}{15}=2\rightarrow z=15\cdot2=30\)

=>Vậy:\(x=8;y=24;z=30\)

Answer 2

[Hồ Thu Giang]

a, \(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{4011}{4011}=1\)

=> x - 1 = 2005 => x = 2006

=> 3 - y = 2006 => y = -2003

b, 3x = y => \(\frac{x}{1}=\frac{y}{3}\)=> \(\frac{x}{4}=\frac{y}{12}\)

5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)=> \(\frac{y}{12}=\frac{z}{15}\)

=> \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

=> \(\frac{6x}{24}=\frac{7y}{84}=\frac{8z}{120}=\frac{6x+7y+8z}{24+84+120}=\frac{456}{10104}=\frac{19}{421}\)

=> x = 19.4 : 421 = \(\frac{76}{421}\)

=> y = 12.19 : 421 = \(\frac{228}{421}\)

=> z = 15.19 : 421 = \(\frac{285}{421}\)