Ta có: \(\left\{{}\begin{matrix}\left|2y-27\right|^{2023}\ge0\forall y\\\left(3x+10\right)^{2024}\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}\ge0\forall x,y\)

Mà: \(\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}=0\)

nên: \(\left\{{}\begin{matrix}2y-27=0\\3x+10=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{27}{2}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: ...

x.3 - 25 = -16

x.3 = -16 + 25

x.3 = 9

x= 9:3

x=3

Không đăng linh tinh nha bạn.

Ý bạn là đề bài làm như thế nào nhỉ, mình chưa hiểu lắm.

..9 \(\times\) ... = ... ...

19 \(\times\) 3  = 57 

:(

837 phút = 13 giờ 57 phút = 13,95 giờ

X x 10 - X x 5 = 5055

X x (10 - 5) = 5055

X x 5 = 5055

X = 5055 : 5

X = 1011

\(X\times10-X\times5=5055\\X\times(10-5)=5055\\X\times5=5055\\X=5055:5\\X=1011\)

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)