Biểu thức này ko tồn tại cả min lẫn max

\(\dfrac{1}{M}=\dfrac{\sqrt{x}-1}{2\sqrt{x}+4}=\dfrac{-\dfrac{1}{4}\left(2\sqrt{x}+4\right)+\dfrac{\sqrt{x}}{2}}{2\sqrt{x}+4}=-\dfrac{1}{4}+\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\)

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\ge0\)

\(\Rightarrow\dfrac{1}{M}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=0\)

\(\left(\dfrac{x}{y}\right)^2=\dfrac{16}{9}\Rightarrow\dfrac{x}{y}=\dfrac{4}{3}\left(x,y>0\right)\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\)

Đặt \(\dfrac{x}{4}=\dfrac{y}{3}=k\Rightarrow x=4k;y=3k\left(k>0\right)\)

\(x^2+y^2=100\\ \Rightarrow16k^2+9k^2=100\\ \Rightarrow k^2=\dfrac{100}{25}=4\\ \Rightarrow k=2\left(k>0\right)\\ \Rightarrow\left[{}\begin{matrix}x=8\\y=6\end{matrix}\right.\)

=>2x=5,8

hay x=2,9

x=2,9

tham khảo

https://hoc24.vn/cau-hoi/tim-so-nguyen-x-de-x2x1-la-boi-cua-x-2.5196424199921#:~:text=l%C3%BAc%2021%3A44-,x,HT,-%C4%90%C3%BAng%200

 vào đi 

https://hoc24.vn/cau-hoi/tim-so-nguyen-x-de-x2x1-la-boi-cua-x-2.5196424199921#:~:text=l%C3%BAc%2021%3A44-,x,HT,-%C4%90%C3%BAng%200

\(x^5+x^4+x+1=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

\(A=-2\left(4a^2-4a+1\right)+5=5-2\left(2a-1\right)^2\le5\)

\(A_{max}=5\) khi \(a=\dfrac{1}{2}\)

a) Ta có: \(A=-8a^2+8a+3\)

\(=-8\left(a^2-a-\dfrac{3}{8}\right)\)

\(=-8\left(a^2-2\cdot a\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=-8\left(a-\dfrac{1}{2}\right)^2+5\le5\forall a\)

Dấu '=' xảy ra khi \(a=\dfrac{1}{2}\)

\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)

=>x=4