Trước tiên ta cần chứng minh : \(1^2+n^2+\dfrac{n^2}{\left(n+1\right)^2}\text{=}\left(n+1-\dfrac{n}{n+1}\right)^2\)

\(\Leftrightarrow2.\left(\dfrac{n\left(n+1\right)}{n+1}-\dfrac{n}{n+1}-\dfrac{n^2}{n+1}\right)\text{=}0\)

\(\Leftrightarrow2.0\text{=}0\left(LĐ\right)\)

Ta có : \(E\text{=}\sqrt{1+2007^2+\dfrac{2007^2}{2008^2}}+\dfrac{2007}{2008}\)

Với bổ đề trên thì :

\(E\text{=}\sqrt{\left(2007+1-\dfrac{2007}{2008}\right)^2}+\dfrac{2007}{2008}\)

\(E\text{=}2008+\dfrac{2007}{2008}-\dfrac{2007}{2008}\)

\(E\text{=}2008\)

Trước tiên ta cần phải rút gọn biểu thức A trước.

Ta có : \(A\text{=}\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}\)

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

\(A\text{=}\dfrac{\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}}{\sqrt{x+\sqrt{2x+1}+\sqrt{x-\sqrt{2x+1}}}}\)

\(A\text{=}\dfrac{\sqrt{x-1}+1+|\sqrt{x-1}-1|}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\)

\(A\text{=}\dfrac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\left(x\ge2\right)\)

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

\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}}\)

\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}}\)

\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{2x-1}+1+\sqrt{2x-1}-1}\left(x\ge2\right)\)

\(A\text{=}\dfrac{\sqrt{2x-2}}{\sqrt{2x-1}}\)

Xét tử thức và mẫu thức của A ta thấy :

\(\sqrt{2x-2}< \sqrt{2x-1}\left(x\ge2\right)\)

\(\Rightarrow A< 1\left(đpcm\right)\)

\(\dfrac{3}{4}\times\left(\dfrac{5}{6}+\dfrac{7}{8}\right)\text{=}\dfrac{3}{4}\times\left(\dfrac{40}{48}+\dfrac{42}{48}\right)\)

\(\text{=}\dfrac{3}{4}\times\left(\dfrac{41}{24}\right)\)

\(\text{=}\dfrac{41}{32}\)

\(\dfrac{3}{2}:\dfrac{8}{4}+\dfrac{21}{48}\text{=}\dfrac{3}{2}:2+\dfrac{21}{48}\)

\(\text{=}\dfrac{3}{4}+\dfrac{21}{48}\text{=}\dfrac{36}{48}+\dfrac{21}{48}\text{=}\dfrac{57}{48}\)

Có điều kiện là n không vậy bạn.

( 1-1/2) : (1-1/3) : ( 1-1/4) = 1/2 : 2/3 : 3/4

 = \(\dfrac{1}{2}\times\dfrac{3}{2}\times\dfrac{4}{3}\text{=}1\)

\(\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\)

\(\text{=}\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\text{=}\dfrac{5}{2}\)

c) ( 57-725) - ( 605-53)

 = 57-725-605+53

 = ( 57+53) -( 725+605)

 = 110-1330

 = -1220

a) -72.17+72.31 - 36 . 228

 = -72.17 + 72. 31 - 36.2.114

 = -72 .17 + 72 . 31 -72 . 114

 = -72. ( 17-31+114)

 = -72 . 100

 = -7200

b) ..........

 = -7 . ( 16-36.9) -(-125) 

 = -7 . ( 16-324) +125

 =-7 . ( -308) +125

 = 2156+125

 = 2281

\(2^x+1\text{=}2^3\)

\(2^x\text{=}8-1\)

\(2^x\text{=}7\)

Bạn xem lại đề nhé.

Số hạng của dãy số trên là : \(\left(2026-1\right):1+1\text{=}2026\)

Ta xét với cặp : 1-2 ; 3-4 ; ......... ; 2025-2026=-1

Tổng của dãy số trên là : \(\dfrac{\left(1-2\right).2026}{2}\text{=}-1013\)