\(VT=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)

      \(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

      \(=\left|4+\sqrt{7}\right|-\sqrt{7}\)

      \(=4+\sqrt{7}-\sqrt{7}\)

      \(=4=VP\) (đpcm)

tự tính

1+0.1.2.3.....100

= 1 + 0

= 1 

đúng thì kicks nha bạn

1+1x2=3

đúng ko

đúng k nhé

hi

tk namhim27

mk 000000 

ai lay acc ko

\(y=\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}+\sqrt{\frac{x^2}{4}-\sqrt{x^2-4}}\) Điều kiện: \(x\ge2\)

\(\Rightarrow2y=2.\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}+2.\sqrt{\frac{x^2}{4}-\sqrt{x^2-4}}\)

\(=\sqrt{x^2+4\sqrt{x^2-4}}+\sqrt{x^2-4\sqrt{x^2-4}}\)

\(=\sqrt{x^2-4+4\sqrt{x^2-4}+4}+\sqrt{x^2-4-4\sqrt{x^2-4}+4}\)

\(=\sqrt{\left(\sqrt{x^2-4}+2\right)^2}+\sqrt{\left(\sqrt{x^2-4}-2\right)^2}\)

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

\(=\sqrt{x^2-4}+2+\left|\sqrt{x^2-4}-2\right|\)(1)

TH1: \(\sqrt{x^2-4}-2\ge0\Rightarrow\sqrt{x^2-4}\ge2\Rightarrow x^2-4\ge4\Rightarrow x\ge2\sqrt{2}\).Ta có:

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

Do \(x\ge2\sqrt{2}\Rightarrow2\sqrt{x^2-4}\ge2\sqrt{\left(2\sqrt{2}\right)^2-4}=4\)

TH2:  \(\sqrt{x^2-4}-2< 0\Rightarrow\sqrt{x^2-4}< 2\Rightarrow x^2-4< 4\Rightarrow x^2< 8\Rightarrow2\le x< 2\sqrt{2}\).Ta có:

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

Vậy GTNN của y bằng 4.

Dấu "=" xảy ra khi \(2\le x\le2\sqrt{2}\)

\(\forall n\inℕ^∗\)ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)  (*)

Thay n=1; n=2; n=3; .....; n=2004 Ta có:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}\)

\(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

=>  \(\sqrt{2}.A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

=> \(A=\sqrt{2}\)

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

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

\(=\frac{\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|}{\sqrt{2}}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)