a) Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{2+x-y}{4011}=\frac{2+4009}{4011}=1\)

=> \(\begin{cases}x-1=2005\\3-y=2006\end{cases}\)\(\Leftrightarrow\begin{cases}x=2006\\y=-2003\end{cases}\)

b) Có: \(3x=y\Rightarrow\frac{x}{1}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{12}\)

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

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

Áp dụng tc của dãy tỉ số bằng nahu ta có:

\(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{6x+7y+8z}{6\cdot4+7\cdot12+8\cdot15}=\frac{456}{228}=2\)

=> \(\begin{cases}x=8\\y=24\\z=30\end{cases}\)

c) Có: \(x-24=y\Rightarrow x-y=24\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{24}{4}=6\)

=> \(\begin{cases}x=42\\y=18\end{cases}\)

a) x-1/2005=3-y/2006

áp dụng tc dãy ts = nhau ta có :

x-1/2005=3-y/2006=(x-1)+(3-y)/2005+2006=x-1+3-y/4011=x-y-1+3/4001=4009-1+3/4011=4011/4011=1

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

=>3-y/2006=1=>3-y=2006=>y=-2003

vậy...

c) 

3x=y

=>x/1=y/3

=>x/4=y/12

5y=4z

=>y/4=z/5

=>y/12=z/15

=>x/4=y/12=z/15

=>6x/24=7y/84=8z/120

áp dụng tc dãy ts = nhau ta có :

6x/24=7y/84=8z/120 = 6x+7y+8z/24+84+120=456/228=2

=>x/4=2=>x=8

=>y/12=2=>y=24

=>z/15=2=>z=30

vậy ...

Tìm x,y,z biết:

a) x−12005 =3−y2006 và x-y=4009

a)x-1=4009

x=4009+1

x=4010

=>\(\frac{4010}{2005}\)

3-y=4009

y=3-4009

y=-4006

=>\(\frac{-4006}{2006}\)

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:\(x=8;y=24;z=30\)

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 = \(\frac{285}{421}\)

1,x/7=y/3 va x-24=y

=>x/7=y/3 va x-y=24

adtcdts=n: 

x/7=y/3=x-y/7-3=24/4=6

Suy ra :x/7=6=>x=6.742

y/3=6=>y=3.6=18

2,Adtcdts=n:

x/5=y/7=z/2=y-x/7-5=48/2=24

suy ra : x/5=24=>x=120

y/7=24=>y=168

z/2=24=>z=48

a) \(\frac{x}{1}=\frac{y}{3}=\frac{4z}{15}=\frac{6x+7y+8z}{1.6+3.7+15.2}=\frac{456}{57}=8\)

x=8

y=24

z=30

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

hay  \(\frac{x}{4}=\frac{y}{12}\)

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

hay  \(\frac{y}{12}=\frac{z}{15}\)

suy ra:   \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

đến đây bạn ADTCDTSBN nhé

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

hay  \(\frac{x}{4}=\frac{y}{12}\)

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

hay  \(\frac{y}{12}=\frac{z}{15}\)

suy ra:   \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

đến đây bạn ADTCDTSBN nhé

\(\Rightarrow\frac{x}{1}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{6x+7y+8z}{6.4+7.12+8.15}=\frac{456}{228}=2\)

=> x= 4.2 =8

y = 12.2 =24

z = 15.2 =30