Question

(0,4-1,3)²=5,29

(x-3)¹⁰=(x-3)³⁰

(x+1,5)⁸+(2,7-y)¹²=0

Answer 1

a, đề sai

b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)

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

Vậy .....

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{12}=0\)

Vì: \(\left\{{}\begin{matrix}\left(x+1,5\right)^8\ge0\forall x\\\left(2,7-y\right)^{12}\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\) để bt = 0

=>\(\left\{{}\begin{matrix}\left(x+1,5\right)^8=0\\\left(2,7-y\right)^{12}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

Vậy.............