Question

1.Tìm x,y 

\(|9y-1|+\left(2x+3\right)^2=0\)

2.So sánh 

a) \([\left(\frac{-1}{3}\right)^7]^4\)và \([\left(\frac{-1}{2}\right)^{14}]^2\)

b)\((\frac{-2}{3})^{12}\)và \(\left(\frac{4}{9}\right)^7\)

Answer 1

1) Ta có: \(\left|9y-1\right|+\left(2x+3\right)^2=0\)

Mà \(\hept{\begin{cases}\left|9y-1\right|\ge0\\\left(2x+3\right)^2\ge0\end{cases}}\left(\forall x,y\right)\)

=> \(\left|9y-1\right|+\left(2x+3\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|9y-1\right|=0\\\left(2x+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}9y-1=0\\2x+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

Answer 2

2)

a) Ta có: \(\left[\left(-\frac{1}{3}\right)^7\right]^4=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)

và \(\left[\left(-\frac{1}{2}\right)^{14}\right]^2=\left(\frac{1}{2}\right)^{28}=\frac{1}{2^{28}}\)

Vì \(\frac{1}{3^{28}}< \frac{1}{2^{28}}\Rightarrow\left[\left(-\frac{1}{3}\right)^7\right]^4< \left[\left(-\frac{1}{2}\right)^{14}\right]^2\)

b) Ta có: \(\left(-\frac{2}{3}\right)^{12}=\left[\left(-\frac{2}{3}\right)^2\right]^6=\left(\frac{4}{9}\right)^6\)

Ta thấy \(0< \frac{4}{9}< 1\)\(\Rightarrow\left(\frac{4}{9}\right)^6>\left(\frac{4}{9}\right)^7\)

\(\Rightarrow\left(-\frac{2}{3}\right)^{12}>\left(\frac{4}{9}\right)^7\)