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1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)(đk x+y+z\(\ne0\)
\(\Rightarrow\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=0,5\)
\(\Rightarrow y+z=0,5-x,x+z=0,5-y,x+y=0,5-z\)
\(\Rightarrow\frac{0,5-x+1}{x}=2\Rightarrow\frac{1,5-x}{x}=2\Rightarrow1,5-x=2x\Rightarrow3x=1,5\Rightarrow x=\frac{1}{2}\)
\(\Rightarrow\frac{0,5-y+2}{y}=2\Rightarrow\frac{2,5-y}{y}=2\Rightarrow2,5-y=2y\Rightarrow3y=2,5\Rightarrow y=\frac{5}{6}\)
\(\Rightarrow z=0,5-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)
Vậy \(x=\frac{1}{2},y=\frac{5}{6},z=-\frac{5}{6}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\)\(\frac{x+y-3}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\frac{y+z+1}{x}+1=\frac{\frac{3}{2}}{x}=3\Rightarrow x=\frac{1}{2}\)
Tương tự suy ra \(y=\frac{5}{6},z=-\frac{5}{6}\)
k cho mình nha!
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{1}{x+y+z}=\frac{x+y-3+y+z+1+x+z+2}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
Xét \(\frac{x+y-3}{z}=2\)
\(\Rightarrow x+y-3=2z\)
\(\Rightarrow x+y+z-3=3z\)
\(\Rightarrow\frac{1}{2}-3=3z\)
\(\Rightarrow\frac{-5}{2}=3z\)
\(\Rightarrow z=\frac{-5}{6}\)
Xét \(\frac{y+z+1}{x}=2\)
\(\Rightarrow y+z+1=2x\)
\(\Rightarrow x+y+z+1=3x\)
\(\Rightarrow\frac{1}{2}+1=3x\)
\(\Rightarrow\frac{3}{2}=3x\)
\(\Rightarrow x=\frac{1}{2}\)
Xét \(\frac{x+z+2}{y}=2\)
\(\Rightarrow x+z+2=2y\)
\(\Rightarrow x+y+z+2=3y\)
\(\Rightarrow\frac{1}{2}+2=3y\)
\(\Rightarrow\frac{5}{2}=3y\)
\(\Rightarrow y=\frac{5}{6}\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{2};\frac{5}{6};\frac{-5}{6}\right)\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{z+y+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
=\(\frac{z+y+1+x+z+2+x+y-3}{x+y+z}\)
= \(\frac{2\left(x+y+z\right)}{x+y+z}=2\)(Do x + y + z \(\ne\)0)
=> \(\frac{1}{x+y+z}=2\) => x + y+ z = 1/2
=> \(\frac{z+y+1}{x}=2\) => \(z+y+1=2x\) => z + y + x = 3x - 1 => 3x - 1 = 1/2 => 3x = 3/2 => x = 1/2
=> \(\frac{x+z+2}{y}=2\) => x + z + 2 = 2y => x + y + z + 2 = 3y => 3y = 5/2 => y = 5/6
=> \(\frac{x+y-3}{z}=2\) => x +y - 3 = 2z => x + y + z - 3 = 3z => 3z = -5/2 => z = -5/6
Vậy ...
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{z+y+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=\frac{2}{1}\)
hay \(\frac{x+y+z}{1}=\frac{1}{2}=0,5\)
\(\Rightarrow x+y+z=0,5\)
\(\Rightarrow y+z=0,5-x\)
\(x+z=0,5-y\)
\(x+y=0,5-z\)
+ Ta có :
\(\frac{z+y+1}{x}=\frac{0,5-x+1}{x}=2\)
\(\Rightarrow1,5-x=2x\)
\(3x=1,5\)
\(x=0,5\)
+ Ta có : \(\frac{x+z+2}{y}=\frac{0,5-y+2}{y}=2\)
\(\Rightarrow2,5-y=2y\)
\(3y=2,5\)
\(y=\frac{5}{6}\)
+ Ta có :
\(\frac{x+y-3}{z}=\frac{0,5-z-3}{z}=2\)
\(\Rightarrow-2,5-z=2z\)
\(3z=-2,5\)
\(z=-\frac{5}{6}\)
Vậy \(x=0,5;y=\frac{5}{6};z=-\frac{5}{6}\)
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{z+y-3}{z}=\frac{1}{x+y+z}\)
\(=\frac{y+z+z+x+x+y+1+2-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\frac{y+z+1}{x}=2\)
\(\Rightarrow y+z+1=2x\)
\(x+y+z+1=3x\Rightarrow\frac{3}{2}=3x\)
Tương tự với mấy cái khác bạn tính được x,y,z
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+z+x+2+x+y-3}{x+y+z}\)
\(\Rightarrow\frac{1}{x+y+z}=\frac{2x+2y+2z}{x+y+z}\)
\(\Rightarrow1=2\left(x+y+z\right)\)
\(\Rightarrow x+y+z=\frac{1}{2}\left(1\right)\)
Thay vào đề đc :
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{\frac{1}{2}}=2\)
\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(2\right)\\z+x+2=2y\left(3\right)\\x+y-3=2z\left(4\right)\end{cases}}\)
Từ (2) => x + y + z + 1 = 3x
Thay (1) vào đc \(\frac{1}{2}+1=3x\)
\(\Leftrightarrow3x=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{1}{2}\)
Từ (3) => x + y + z + 2 = 3y
Thay (1) vào đc \(\frac{1}{2}+2=3y\)
\(\Leftrightarrow y=\frac{5}{6}\)
Khi đó \(z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)
Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)
\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{2x+2y+2x}{x+y+z}\)=\(\frac{1}{x+y+z}\)
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
Đặt \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=k\)
Áp dụng TC DTSBN ta có : \(k=\frac{2\left(x+y+z\right)+1+2-3}{x+y+z}=2\)
\(\Rightarrow y+z+1=2x;x+z+2=2y;x+y-3=2z;x+y+z=\frac{1}{2}\)
Từ \(y+z+1=2x\Leftrightarrow x+y+z+1=3x\Leftrightarrow\frac{1}{2}+1=3x\Rightarrow x=\frac{1}{2}\)
Từ \(x+z+2=2y\Leftrightarrow x+y+z+2=3y\Leftrightarrow\frac{1}{2}+2=3y\Rightarrow y=\frac{5}{6}\)
Từ \(x+y-3=2z\Leftrightarrow x+y+z-3=3z\Leftrightarrow\frac{1}{2}-3=3z\Rightarrow z=-\frac{5}{6}\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}\)