tìm x , y ,z thuộc Q , biết x+y = \(\frac{1}{2}\), y+z = \(\frac{1}{3}\), x +z = \(\frac{1}{4}\)
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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 \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
a)ta có xy=7*9=7*3*3
vậy x =9;21 , y=7;3
b) xy=-2*5
mà x<0<y
nên x=-2 ,y=5
c)x-y=5 hay x=y+5
\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\Leftrightarrow\left(x+y\right)\left(\frac{zx+z^2+zy+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=0\).
Vậy \(M=\frac{3}{4}+\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=\frac{3}{4}+0=\frac{3}{4}\)
\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)
\(\frac{2x+2y+2x}{x+y+z}\)=\(\frac{1}{x+y+z}\)
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
Ta có : x+y=1/2=>x=1/2-y
y+z=1/3=>z=1/3-y
=>x-z=1/2-y-1/3+y=1/2-1/3-(y-y)=1/6
Vậy x = (1/6+1/4):2=5/24
z = (1/4-1/6):2=1/24
=> y = 1/3-1/24=7/24
Ta có: \(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
\(x+y+z=\frac{13}{12}:2=\frac{13}{24}\)
\(x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)
\(y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)
\(z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)
Vậy x = ....; y = .....; z = .......
k cho mik nha