ĐKXĐ: \(x\ge0\)

Áp dụng BĐT Cauchy cho 2 số dương: \(x+4\ge2\sqrt{4x}=4\sqrt{x}\)

\(\Rightarrow\frac{4\sqrt{x}}{x+4}\le1\Leftrightarrow T\le1\)

Dấu "=" xảy ra \(\Leftrightarrow x=4\)

ĐKXĐ: \(x\ge0,x\ne1\)

\(F=\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\)\(\sqrt{x}+1\)và \(\sqrt{x}-1\)trái dấu 

\(\Rightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Vậy \(0< x< 1\)thỏa mãn đề bài.

Ta có: \(\left(x+x^2\right)^{2+1}=0\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]^3=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

a) đk: \(x\ge0\)

Ta có: \(M< -\frac{1}{2}\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+3}+\frac{1}{2}< 0\)

\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)

Mà \(2\left(\sqrt{x}+3\right)>0\left(\forall x\right)\Rightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Vậy \(0\le x< 9\)

b) Ta có: \(M=\frac{-3}{\sqrt{x}+3}\ge-\frac{3}{0+3}=-1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Min(M) = -1 khi x = 0

Ta có : 8x\(\le\)48

     <=>  x\(\le\)48:8

             x\(\le\)6

Chia hai vế cho 8, giữ nguyên chiều bất đẳng thức và kết luận.

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