Question

[Nguyễn Tuấn Anh]

giúp mình với

Answer 1

[Yen Nhi]

Answer:

\(7x\left(4x-\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x=0\\4x-\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{8}\end{cases}}\)

\(\left(2x-1\right)\left(1,5x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\1,5x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=1\\1,5x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{3}\end{cases}}\)

\(\left(2x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow2x-\frac{1}{2}=0\)

\(\Rightarrow2x=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{4}\)

\(\left(x-2\right)^2+\left|y^2-9\right|=0\)

Có: \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left|y^2-9\right|\ge0\forall y\end{cases}}\) mà đề ra \(\left(x-2\right)^2+\left|y^2-9\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left|y^2-9\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x-2=0\\y^2-9=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y^2=9\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y\in\pm3\end{cases}}\)