Ta có: \(\dfrac{x}{3}=\dfrac{y}{6}\Rightarrow x=\dfrac{3y}{6}=\dfrac{1}{2}y\)

Theo đề bài ta có : \(xy=162\Rightarrow\dfrac{1}{2}y.y=162\Rightarrow y^2=324\Rightarrow y=18\)

\(\Rightarrow x=\dfrac{1}{2}y=9\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=6k\end{matrix}\right.\)

Ta có: xy=162

\(\Leftrightarrow18k^2=162\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=9\\y=6k=18\end{matrix}\right.\)

Trường hợp 2: k=-3

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-9\\y=6k=-18\end{matrix}\right.\)

y + y x 2,7 + y x 6,3 = 162 : 15

   y x (1 + 2,7 + 6,3) = 10,8

                      y x 10 = 10,8

                              y = 10,8 : 10

                              y = 1,08

162 / 57 x y = 270 / 335

               y  = 270 / 335 : 162 / 57

               y  = 19 / 67

Vậy , y = 19 / 67

\(\frac{162}{57}\)x y = \(\frac{270}{335}\)

y = \(\frac{270}{335}\): \(\frac{162}{57}\)

y = \(\frac{19}{67}\)

Chúc bạn học tốt!

a/Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{x+y}{3+6}=\dfrac{90}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10\cdot3=30\\y=10\cdot6=60\end{matrix}\right.\)
Vậy ...
b/Ta có:
\(\dfrac{x}{3}=\dfrac{4x}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{4x}{12}=\dfrac{y}{6}=\dfrac{4x-y}{12-6}=\dfrac{42}{6}=7\)
\(\Rightarrow\left\{{}\begin{matrix}x=7\cdot3=21\\y=7\cdot6=42\end{matrix}\right.\)
Vậy ...
c/Đặt \(x=k;y=k\) ( k \(\in\) N* )
\(\Rightarrow x=3k;=6k\)
Mà \(xy=162\)
\(\Rightarrow3k\cdot6k=162\)
\(\Rightarrow18k^2=162\)
\(\Rightarrow k^2=9\)
\(\Rightarrow k=\pm3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\cdot3=9\\x=\left(-3\right)\cdot3=-9\\y=3\cdot6=18\\y=\left(-3\right)\cdot6=-18\end{matrix}\right.\)
Vậy ...
#NoSimp  

a)\(162:y+216:y=6\)

⇔\(\left(162+216\right):y=6\)

⇔\(378:y=6\)

⇔\(y=63\)

b)\(602:y+123.2=253\)

⇔\(602:y+246=253\)

⇔\(602:y=7\)

⇔\(y=86\)

c)\(8\left(15720-y\right)=51784\)

⇔\(125760-8y=51784\)

⇔\(8y=73976\)

⇔\(y=9247\)

x=72

y=60

z=30

Vì \(\left(x-5\right)^{2018}\ge0;\left|2y^2-162\right|^{2018}\ge0\Rightarrow\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}\ge0\)

mà \(\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}=0\)

Dấu ''='' xảy ra khi x = 5 ; \(2y^2=162\Leftrightarrow y^2=81\Leftrightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)

Vì \(\left(x-5\right)^{2018}\ge0\\ \left|2y^2-162\right|^{2018}\ge0\\ \)

Suy ra phương trình dc thỏa mãn khi và chỉ khi x-5 = 0 và 2y^2-162=0

\(\left\{{}\begin{matrix}\left(x-5\right)^{2018}=0\\\left|2y^2-162\right|^{2018}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\2\left(y^2-81\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\pm9\end{matrix}\right.\)