Đặt \(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}=x>0\)

\(\Rightarrow x^3=14-3\left(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}\right)\sqrt[3]{\left(5\sqrt[]{2}+7\right)\left(5\sqrt[]{2}-7\right)}\)

\(\Rightarrow x^3=14-3x.\sqrt[3]{\left(5\sqrt[]{2}\right)^2-7^2}\)

\(\Rightarrow x^3=14-3x\)

\(\Rightarrow x^3+3x-14=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)=0\)

\(\Rightarrow x=2\)

\(VT=2\sqrt{2+\sqrt{3}}=\sqrt{2}.\sqrt{3+2\sqrt{3}+1}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}=VP\) Vậy , đẳng thức được chứng minh .

BĐVT có :\(2\sqrt{2+\sqrt{3}}=\sqrt{2}.\sqrt{3+2\sqrt{3}+1}\)

=\(\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+2\)

\(\Rightarrow\)VT=VP(đpcm)

ĐK:`x>=0`

Nhân hai vế với `sqrt{x+1}`

`2sqrt2+sqrt{x^2+x}<=sqrt{x^2+10x+9}`

BP 2 vế ta có:

`8+x^2+x+4\sqrt{2x^2+2x}<=x^2+10x+9`

`<=>4\sqrt{2x^2+2x}<=9x-1`

ĐK:`x>=1/9`

`<=>16(2x^2+2x)<=81x^2-18x+1`

`<=>32x^2+32x<=81x^2-18x+1`

`<=>49x^2-50x+1>=0`

`<=>(x-1)(49x-1)>=0`

Vì `x>=1/9=>49x-1>0`

`=>x-1>=0<=>x>=1`

Vậy bpt có nghiệm `S={x|x>=1}`

Đề bài đúng: \(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)

Hoặc: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=1\)

\(=\dfrac{\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

\(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

Ta có:

\(\dfrac{\sqrt[4]{17+12\sqrt{2}} +\sqrt[4]{17-12\sqrt{2}}}{2}\)

\(=\dfrac{\sqrt[4]{3^2+2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}+\sqrt[4]{3^2-2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt[4]{\left(3+2\sqrt{2}\right)^2}+\sqrt[4]{\left(3-2\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt[4]{\left(2+2\sqrt{2}+1\right)^2}+\sqrt[4]{\left(2-2\sqrt{2}+1\right)^2}}{2}\)

\(=\dfrac{\sqrt[4]{[\left(\sqrt{2}+1\right)^2]^2}+\sqrt[4]{[\left(\sqrt{2}-1\right)^2]^2}}{2}\)

\(=\dfrac{\sqrt[4]{\left(\sqrt{2}+1\right)^4}+\sqrt[4]{\left(\sqrt{2}-1\right)^4}}{2}\)

\(=\dfrac{\sqrt{2}+1+\sqrt{2}-1}{2}\)

\(=\dfrac{2\sqrt{2}}{2}\)

\(=\sqrt{2}\) (đpcm)

Ta có :

\(\hept{\begin{cases}\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\end{cases}}\forall n\in N\)

Suy ra : \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)

Đặt \(M=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2499}}+\frac{1}{\sqrt{2500}}\)

\(\Leftrightarrow\frac{1}{2}M=\frac{1}{2\sqrt{2500}}+\frac{1}{2\sqrt{2499}}+...+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{2}}+\frac{1}{2}\)

Áp dụng BĐT , ta có :

\(\frac{1}{2}M< \sqrt{2500}-\sqrt{2499}+\sqrt{2499}-\sqrt{2498}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}M< \sqrt{2500}-\sqrt{1}+\frac{1}{2}=50-\frac{1}{2}< 50\)

\(\Rightarrow M< 100\)