Lời giải:
Đặt biểu thức đã cho là $P$

\(2P=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+....+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}(*)\)

Mà:

\(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+....+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{(\sqrt{1}+\sqrt{3})(\sqrt{3}-\sqrt{1})}+\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3})}+....+\frac{\sqrt{101}-\sqrt{99}}{(\sqrt{99}+\sqrt{101})(\sqrt{101}-\sqrt{99})}\)

\(=\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+...+\frac{\sqrt{101}-\sqrt{99}}{2}\)

\(=\frac{\sqrt{101}-\sqrt{1}}{2}>\frac{\sqrt{100}-1}{2}=\frac{9}{2}(**)\)

Từ \((*); (**)\Rightarrow 2P>\frac{9}{2}\Rightarrow P>\frac{9}{4}\) (đpcm)

Ta có :

\(2\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\right)\)

\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\frac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\frac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\frac{9}{2}\)

\(\Rightarrow\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}>\frac{9}{4}\left(đpcm\right)\)

Chúc bạn học tốt !!!

Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{3}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Rightarrow2A=\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{1}\right)^2}{\sqrt{1}+\sqrt{3}}+...+\frac{\left(\sqrt{99}\right)^2-\left(\sqrt{97}\right)^2}{\sqrt{97}+\sqrt{99}}\)

\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{99}-\sqrt{97}\)

\(=\sqrt{99}-1\)

\(\Rightarrow A=\frac{\sqrt{99}-1}{2}=\frac{2\sqrt{99}-2}{4}>\frac{9}{4}\left(đpcm\right)\)

Gọi \(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Rightarrow2A=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}\)

\(=\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{1}\right)^2}{\sqrt{3}+\sqrt{1}}+...+\frac{\left(\sqrt{99}\right)^2-\left(\sqrt{97}\right)^2}{\sqrt{99}+\sqrt{97}}\)

\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{99}-\sqrt{97}\)

\(=\sqrt{99}-1\)

Vậy \(A=\frac{\sqrt{99}-1}{2}=\frac{2\sqrt{99}-2}{4}>\frac{9}{4}\)

\(2A=\frac{2}{1+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}\)

\(2A>\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

Nhân liên hợp tử - mẫu vế phải:

\(\Rightarrow2A>\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(2A>\frac{1}{2}\left(\sqrt{101}-1\right)>\frac{1}{2}\left(\sqrt{100}-1\right)=\frac{9}{2}\)

\(\Rightarrow A>\frac{9}{4}\)

Đặt:

\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Leftrightarrow2A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

\(=\frac{1}{2}.\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(=\frac{1}{2}.\left(\sqrt{101}-\sqrt{1}\right)>\frac{1}{2}.\left(\sqrt{100}-\sqrt{1}\right)\)

\(=\frac{9}{2}\)

\(\Rightarrow A>\frac{9}{4}\)

Câu 2/ Ta có:

\(n^{n+1}>\left(n+1\right)^n\)

\(\Leftrightarrow n>\left(1+\frac{1}{n}\right)^n\left(1\right)\)

Giờ ta chứng minh cái (1) đúng với mọi \(n\ge3\)

Với \(n=3\) thì dễ thấy (1) đúng.

Giả sử (1) đúng đến \(n=k\) hay

\(k>\left(1+\frac{1}{k}\right)^k\)

Ta cần chứng minh (1) đúng với \(n=k+1\)hay \(k+1>\left(1+\frac{1}{k+1}\right)^{k+1}\)

Ta có: \(\left(1+\frac{1}{k+1}\right)^{k+1}< \left(1+\frac{1}{k}\right)^{k+1}=\left(1+\frac{1}{k}\right)^k.\left(1+\frac{1}{k}\right)\)

\(< k\left(1+\frac{1}{k}\right)=k+1\)

Vậy có ĐPCM

bằng 122223

Bạn trục căn thức ở mẫu rồi trừ đi là xong nhé,vì khi trục căn thức thì ở A mẫu chung là 1,ở B mẫu chung là 2.

giai ra giup mik di

A=(√3-√2)/(3-2)+(√4-√3)/(4-3)+......

=√3-√2+√4-√3+......+√25-√24

=√25-√2=5-√2.Câu b tương tự

bài này dễ mà bạn cần mk giải chi tiết ko

kết quảA =\(\frac{\sqrt{99}-\sqrt{3}}{2}\)

Có: \(\frac{1}{\sqrt{n}+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}-\sqrt{n}\right)\left(\sqrt{n+2}+\sqrt{n}\right)}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
\(\Rightarrow A=\frac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{99}-\sqrt{97}}{2}\)
\(A=\frac{\sqrt{99}-\sqrt{3}}{2}\)

a.

\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}\)

\(=\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{9}-\sqrt{7}}{9-7}\)

\(=\frac{\sqrt{9}-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{1}}{2}\)

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

b.

\(B=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

c.

\(C=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)

\(=\frac{15\sqrt{6}-15}{6-1}+\frac{4\sqrt{6}+8}{6-4}-\frac{36+12\sqrt{6}}{9-6}-\sqrt{6}\)

\(=\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}-\sqrt{6}\)

\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)

\(=-11\)

d)D=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)( \(x\ge2\))

=\(\sqrt{x+2\sqrt{2}.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}\)

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

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

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

TH1: \(2\le x\le4\)

Từ (1)<=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}\)

=\(2\sqrt{2}\)

TH2. x\(>4\)

Từ (1) <=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{2}+\sqrt{x-2}\)=\(2\sqrt{x-2}\)

Vậy \(\left[{}\begin{matrix}2\le x\le4\\x>4\end{matrix}\right.< =>\left[{}\begin{matrix}D=2\sqrt{2}\\D=2\sqrt{x-2}\end{matrix}\right.\)

\(=\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{\sqrt{97}-\sqrt{99}}{97-99}\)

\(=\frac{\sqrt{3}-\sqrt{5}+\sqrt{5}-\sqrt{7}+...+\sqrt{97}-\sqrt{99}}{-2}\)

\(=\frac{\sqrt{3}-\sqrt{99}}{-2}=\frac{\sqrt{99}-\sqrt{3}}{2}\)

= \(\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{\sqrt{97}-\sqrt{99}}{97-99}\) = \(\frac{-1}{2}.\left(\sqrt{3}-\sqrt{5}+\sqrt{5}-\sqrt{7}+...+\sqrt{97}-\sqrt{99}\right)\)

= \(-\frac{1}{2}.\left(\sqrt{3}-\sqrt{99}\right)\) = \(\frac{3\sqrt{11}-\sqrt{3}}{2}\)