Tìm x,y,z biết
\(\frac{x+y+2005}{z}=\frac{y+z-2006}{x}=\frac{z+x+1}{y}=\frac{2}{x+y+z}\)
\(\frac{x+y+2005}{z}=\frac{y+z-2006}{x}=\frac{z+x+1}{y}=\frac{2}{x+y+z}\). tìm x,y,z
Tìm X,Y,Z biết:
\(\frac{x+y+2005}{z}=\frac{y+2-2006}{x}=\frac{z+x+1}{y}=\frac{2}{x+y+z}\)
Áp dụng dãy tỉ số bằng nhau
a)\(\frac{x-1}{2005}=\frac{3-y}{2006}\)và x-y=4009
b)\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)và x-y-z=28
Tìm x,y,z khi:
1,\(\frac{x}{7}=\frac{y}{3}vàx-24=y\)
2,\(\frac{x}{5}=\frac{y}{7}=\frac{z}{2}và,y-x=48\)
3,\(\frac{x-1}{2005}=\frac{3-y}{2006}và,x-y=4009\)
4,\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}vã-y-z=28\)
5,\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}và2x+3y-z=-14\)
6,\(3x=y;5y=4zvà6x+7y+8z\)
Tìm x,y,z khi:
a) \(\frac{x-1}{2005}=\frac{3-y}{2006}\) và x-y=4009
b) \(\frac{x}{2}=\frac{y}{3}:\frac{y}{4}=\frac{z}{5}\) và x-y-z=28
c) 3x=y; 5y=4z và 6x+7y+8z=456
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
Tìm x,y,z biết:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-z}{z}=\frac{1}{x+y+z}\)
y+x+z bằng bao nhiêu mới tính ra được chứ?? sai đề à??
Tìm x, y, z biết: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{2x+2y+2x}{x+y+z}\)=\(\frac{1}{x+y+z}\)
2=\(\frac{1}{x+y+z}\)(1)
Từ(1) => \(\frac{1}{x+y+z}\)=2 => x+y+z=0,5=>x+z=0,5-y(2)
Từ(1)=> x+y+1=2x(3)
x+z+2=2y(4)
z+y-3=2z(5)
Thay(2) vào (4) ta được: 0,5-y+2=2y
=> 2,5=3y
=> y=\(\frac{5}{6}\)
Thay y=\(\frac{5}{6}\)vào(3) ta được:x+\(\frac{5}{6}\)+1=2x
\(\frac{11}{6}\)=x
Thay x=\(\frac{11}{6}\); y=\(\frac{5}{6}\)vào x+y+z=0,5 ta đươc:
\(\frac{11}{6}\)+\(\frac{5}{6}\)+z=0,5
z=\(\frac{-13}{6}\)
Vậy ............
chúc bn học tốt.
k cho mik nha
tìm x,y,z biết : \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tìm x y z biết
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow y+z=\frac{1}{2}-x;x+z=\frac{1}{2}-y;z+y=\frac{1}{2}-x\)
THAY VÀO BIỂU THỨC TA CÓ:
\(\frac{\frac{1}{2}-x+1}{x}=2\Rightarrow\frac{3}{2}-x=2x\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\Rightarrow\frac{5}{2}-y=2y\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\Rightarrow\frac{-5}{2}-z=2z\Rightarrow z=-\frac{5}{6}\)
\(\frac{y+z+1}{x}+\frac{x+z+2}{y}+\frac{x+y-3}{z}=\frac{y+x+1+x+z+2+x+y-3}{x+y+x}=\frac{2x+2y+2z}{x+y+z}=2.\)
\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}=0,5\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)\(\Rightarrow\frac{y+z+1}{x}+1=\frac{x+z+2}{y}+1=\frac{x+y-3}{z}+1=0,5+1\)
\(\Leftrightarrow\frac{x+y+z+1}{x}=\frac{x+y+z+2}{y}=\frac{x+y+z-3}{z}=1,5\)
\(\Leftrightarrow\frac{0,5+1}{x}=\frac{0,5+2}{y}=\frac{0,5-3}{z}=1,5\)
\(\Rightarrow\hept{\begin{cases}\frac{1,5}{x}=1,5\\\frac{2,5}{y}=1,5\\\frac{-2,5}{z}=1,5\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=1,6\\z=-1,6\end{cases}}}\)