\(=-10\)

\(=-6\)

câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

Từ \(abc\le\left(\frac{a+b+c}{3}\right)^3\Rightarrow a+b+c\ge3\)

Ta có BĐT phụ \(\frac{1}{2a+1}\ge-\frac{2}{9}a+\frac{5}{9}\)

\(\Leftrightarrow\frac{4\left(a-1\right)^2}{9\left(2a+1\right)}\ge0\forall0< a< 3\) (đúng)

Tương tự ta cũng có:

\(\frac{1}{2b+1}\ge-\frac{2}{9}b+\frac{5}{9};\frac{1}{2c+1}\ge-\frac{2}{9}c+\frac{5}{9}\)

Cộng theo vế 3BĐT trên ta có:

\(VT\ge-\frac{2}{9}\left(a+b+c\right)+\frac{5}{9}\cdot3=1=VP\)

Khi a=b=c=1

lộn đề >=1 nha

Từ 1 đến 10 000 000 thì có : 10 000 000 số

từ 1 dến 10000000 có 10000000 chữ số