\(x\times\left(x\times\frac{1}{2}\right)< 0\)

\(x.\left(x.\frac{1}{2}\right)< 0< =>x^2.\frac{1}{2}< 0< =>x^2< 0< =>x< 0\)

Đề chỉ  thế thôi à

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)

\(=\frac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=\frac{2^{40}}{2^{30}}=\frac{2^{30}.2^{10}}{2^{30}}=2^{10}=1024\)

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=1024\)

\(\left(x-2\right).\left(x+\frac{2}{3}\right)>0\) <=> \(x-2\) và \(x+\frac{2}{3}\) cùng dấu

Xét 2 TH sau:

\(\left(+\right)\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}=>\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}=>x< -\frac{2}{3}}\)

\(\left(+\right)\hept{\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}}=>\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}=>x>2}\)

Vậy \(x< -\frac{2}{3}\) hoặc \(x>2\) thì thỏa mãn......

Ta có: \(2^{30}=\left(2^3\right)^{10}=8^{10}\)     và      \(3^{20}=\left(3^2\right)^{10}=9^{10}\)

Vì \(8^{10}< 9^{10}\)

Vậy  \(2^{30}< 3^{20}\)

A đạt GtLN <=> |x-2|+3 đạt GTNN

Vì  \(\left|x-2\right|\ge0=>\left|x-2\right|+3\ge3\) (với mọi x) =>GTNN của |x-2|+3 là 3

\(=>A=\frac{1}{\left|x-2\right|+3}\le\frac{1}{3}\)

Dấu "=" xảy ra \(< =>\left|x-2\right|=0< =>x=2\)

Vậy \(MaxA=\frac{1}{3}\) khi x=2

\(\frac{8^{13}}{4^{10}}=524288\)

\(\frac{8^{13}}{4^{10}}\)

\(=\frac{\left(2^3\right)^{13}}{\left(2^2\right)^{10}}\)

\(=\frac{2^{3.13}}{2^{2.10}}\)

\(=\frac{2^{39}}{2^{20}}\)

ngồi giãn ước