=>xy/3.4=12/12=1

vậy x/3=1=>x=3

y/4=1=>y=4

2x=6y

=>x=3y

\(x\cdot y=108\)

=>\(3y\cdot y=108\)

=>\(y^2=36\)

=>\(\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\cdot6=18\\x=3\cdot\left(-6\right)=-18\end{matrix}\right.\)

Áp dụng tc dtsbn:

\(\dfrac{x}{y}=\dfrac{5}{3}\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{64}{8}=8\\ \Leftrightarrow\left\{{}\begin{matrix}x=40\\y=24\end{matrix}\right.\)

\(5x=6y\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{5}\)

\(\Rightarrow\dfrac{2x}{12}=\dfrac{3y}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\dfrac{2x}{12}=\dfrac{3y}{15}=\dfrac{2x+3y}{12+15}=\dfrac{7}{27}\)

\(\Rightarrow\dfrac{2x}{12}=\dfrac{7}{27}\)\(\Rightarrow2x=\dfrac{7.12}{27}=\dfrac{28}{9}\)\(\Rightarrow x=\dfrac{14}{9}\)

\(\dfrac{3y}{15}=\dfrac{7}{27}\Rightarrow3y=\dfrac{7.15}{27}=\dfrac{35}{9}\)\(\Rightarrow y=\dfrac{35}{27}\)

\(\frac{4}{5x}=\frac{7}{6y}=\frac{2}{3z}\)

=> \(\frac{5x}{4}=\frac{6y}{7}=\frac{3z}{2}\)

=> \(\frac{5x}{4}:30=\frac{6y}{7}:30=\frac{3z}{2}:30\)

=> \(\frac{x}{24}=\frac{y}{35}=\frac{z}{20}=\frac{x+y+z}{24+35+20}=\frac{80}{79}\)

=> \(x=\frac{1920}{79};y=\frac{2800}{79};z=\frac{1600}{79}\)

1/

Vì $ƯCLN(x,y)=6$ nên đặt $x=6m, y=6n$ với $m,n$ là số tự nhiên, $m,n$ nguyên tố cùng nhau.

Theo bài ra ta có:

$xy=720$

$\Rightarrow 6m.6n=720$

$\Rightarrow mn=20$

Do $m,n$ nguyên tố cùng nhau nên $(m,n)=(1,20), (4,5), (5,4), (20,1)$
$\Rightarrow (x,y)=(6,120), (24,30), (30,24), (120,60)$

2/

Vì $5x=|x+2|+|2x+1|+|x+3|\geq 0$ nên $x\geq 0$

$\Rightarrow |x+2|=x+2; |2x+1|=2x+1; |x+3|=x+3$. Bài toán trở thành:

$x+2+2x+1+x+3=5x$

$\Rightarrow 4x+6=5x$

$\Rightarrow x=6$ (thỏa mãn)

b, Có : |x| + |x-2| = |x| + |2-x| >= |x+2-x| = 2

Lại co : |x-1| >= 0

=> |x|+|x-1|+|x-2| >= 2

Dấu "=" xảy ra <=> x.(2-x) >= 0 và x-1=0 <=> x=1

Vậy x=1

Tk mk nha