đề bài bị lỗi :(

a) \(\begin{cases}\left(x+2\right)^2\ge0\\\left(y-\frac{1}{5}\right)^2\ge0\end{cases}\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10=-10\)hay \(C\ge-10\)

Dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-\frac{1}{5}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{5}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}}}\)

Vậy GTNN C là -10 khi \(\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}.}\)

b)\(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+5\ge0+5=5\)

\(\Rightarrow\frac{4}{\left(2x-3\right)^2-5}\le\frac{4}{5}\Leftrightarrow D\le\frac{4}{5}\)

Dấu "=" xảy ra khi:

\(\left(2x-3\right)^2=0\Rightarrow2x-3=0\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)

Vậy GTLN D là \(\frac{4}{5}\)khi \(x=\frac{3}{2}.\)

thank bạn nha

M = \(\frac{5-x}{x+2}\) (đk \(x\) ≠ - 2)

M = - 1 + \(\frac{7}{x+2}\)

Để Mmax thì \(\frac{7}{x+2}\) max

Vì \(x\) nguyên nên \(\frac{7}{x+2}\) max khi và chỉ khi \(x\) + 2 = 1

Suy ra \(x\) = -1

Với \(x\) = -1 thì M = - 1 + \(\frac{7}{-1+2}\) = - 1 + \(\frac71\) = - 1 + 7 = 6

Vậy Mmax = 6 khi x = - 1

\(đk:x\ge0\)

ta có P=\(x+\sqrt{x}+1=x+2\cdot\frac{1}{2}\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy P min là 3/4 khi x>= 0