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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:
\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{24}{4}=6\)
=> \(\begin{cases}x=42\\y=18\end{cases}\)
Có: x-1/2005=3-y/2006
=> x-1+3-y/2005+2006
=> x-y+2/4011(tính chất của tỷ lệ thức)
Mà x-y=4009 nên
x-1/2005=3-y/2006=x-1+3-y/2005+2006=x-y+2/4011=1
=> x-1/2005=1=> x=2006
=> 3-y/2006=1=> y=-2013
\(3x=y\)=> \(\frac{x}{1}=\frac{y}{3}\)
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
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 ...
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
\(\Rightarrow\frac{x}{1}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{6x+7y+8z}{6.4+7.12+8.15}=\frac{456}{228}=2\)
=> x= 4.2 =8
y = 12.2 =24
z = 15.2 =30
a)x-1=4009
x=4009+1
x=4010
=>\(\frac{4010}{2005}\)
3-y=4009
y=3-4009
y=-4006
=>\(\frac{-4006}{2006}\)