Lời giải:

$x+y=\frac{x}{y}$

$y(x+y)=x$

$x(y-1)+y^2=0$

$x(y-1)=-y^2$

Nếu $y=1$ thì $x+1=x$ (vô lý). Do đó $y\neq 1$

$\Rightarrow x=\frac{y^2}{1-y}$.

Khi đó:
$x+y=3(x-y)$

$\Leftrightarrow \frac{y^2}{1-y}+y=\frac{3y^2}{1-y}-3y$

$\Leftrightarrow \frac{y^2}{1-y}=2y$

$\Leftrightarrow y(\frac{y}{1-y}-2)=0$. Rõ ràng $y\neq 0$ nên $\frac{y}{1-y}-2=0$

$\Leftrightarrow y=2(1-y)\Leftrightarrow y=\frac{2}{3}$

$x=\frac{y^2}{1-y}=\frac{4}{3}$

5 giờ=..... phút

x+y=3(x-y)\(\Rightarrow\)x+y=3x-3y\(\Rightarrow\)y+3y=3x-x\(\Rightarrow\)4y=2x\(\Rightarrow\)2y=x

\(\Rightarrow\)x : y=2\(\Rightarrow\)x+y=2\(\Rightarrow\)2y+y=2\(\Rightarrow\)3y=2

\(\Rightarrow\)y=\(\frac{2}{3}\);  x=\(\frac{1}{3}\)

À quên , x=\(\frac{4}{3}\)

a) Ta có: \(\left|1-2x\right|+\left|2-3y\right|+\left|3-4z\right|\ge0\)

Mà \(\left|1-2x\right|+\left|2-3y\right|+\left|3-4z\right|=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|=0\\\left|2-3y\right|=0\\\left|3-4z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\2-3y=0\\3-4z=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=1\\3y=2\\4z=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{3}\\z=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{2};y=\dfrac{2}{3};z=\dfrac{3}{4}\)

b)xy=x:y=>y²=1

=>y=1 hoặc y=-1

*)y=1

=>x+1=x

=>0x=-1(L)

*)y=-1

=>x-1=-x

=>2x=1

=>x=1/2

              Vậy y=-1 x=1/2

c)xy=x:y=>y²=1

=>y=1 hoặc y=-1

*)y=1

=>x-1=x

=>0x=1(L)

*)y=-1

=>x+1=-x

=>2x=-1

=>x=-1/2

Vậy y=-1 x=-1/2

d)x(x+y+z)+y(x+y+z)+z(x+y+z)=-5+9+5=9

=>(x+y+z)²=9

=>x+y+z=3 hoặc x+y+z=-3

*)x+y+z=3

=>x=-5:3=-5/3

y=9:3=3

z=5:3=5/3

*)x+y+z=-3

=>x=-5:(-3)=5/3

y=9:(-3)=-3

z=5:(-3)=-5/3