mình cần gấp

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)  \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)

\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)

\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)

Thay vào lần lượt ta có:

\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)

\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)

\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)

1/x + 1/y + 1/z = 1/x+y+z

<=> xy+yz+zx/xyz = 1/x+y+z

<=> (xy+yz+xz).(x+y+z)=xyz

<=> x^2y+xy^2+y^2z+z^2y+z^2x+x^2z+3xyz=xyz

<=> x^2y+y^2x+y^2z+z^2y+z^2x+x^2z+2xyz = 0

<=> (x+y).(y+z).(z+x) = 0

<=> x+y=0 hoặc y+z=0 hoặc x+z=0 

<=> x=-y hoặc y=-z hoặc z=-x

Nếu x=-y => x^25 = -y^25 => P = 0

Nếu y=-z => y^3 = -z^3 => P = 0

Nếu z=-x => z^2006 = x^2006 => P = 0

Vậy P = 0

Tk mk nha

Từ: \(\frac{x-1}{2015}\)=\(\frac{3-y}{2016}\)=\(\frac{x-1+3-y}{2015+2016}\)=\(\frac{x-y+2}{4011}\)=\(\frac{4009+2}{4011}\)=\(\frac{4011}{4011}\)=1

Do:\(\frac{3-y}{2016}\) = 1 => x=2015

                                      x =2015+ 1 

                                       x=2016

\(\frac{3-y}{2016}\)= 1 => y =2016

                              y=3- 2016

                               y= -2013

ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2006}\)    (x;y;z khác 0)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)(vì x+y+z=2006)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-\left(x+y+z\right)}{\left(x+y+z\right).z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{\left(x+y+z\right).z}\)

\(\Leftrightarrow-\left(x+y\right)xy=\left(x+y\right)\left(xz+yz+z^2\right)\)  (vì x;y;z khác 0)

\(\Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=>  x+y=0 hoặc y+z=0 hoặc z+x=0

mà x+y+z=2006 nên

z=2006 hoặc x=2006 hoặc y=2006 

=> đpcm