Cho A = \(\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}\). CM A < 1với a ≥ 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.



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
a)
�)−72.17+72.31−36.228-72.17+72.31-36.228
=−72.17+72.31−36.2.114=-72.17+72.31-36.2.114
=−72.17+72.31−72.114=-72.17+72.31-72.114
=72(31−17−114)=72(31-17-114)
=72.(−100)=−7200=72.(-100)=-7200
b)�)−7.[(−2)4+(−36):(−32)]−(−5)3-7.[(-2)4+(-36):(-32)]-(-5)3
=−7.[16−36:(−9)]+125=-7.[16-36:(-9)]+125
=−7.[16−−4]+125=-7.[16--4]+125
=−7.20+125=-7.20+125
=−140+125=-140+125
=−15=-15
c)�)(57−725)−(605−53)(57-725)-(605-53)
=57−725−605+53=57-725-605+53
=(57+53)−120=(57+53)-120
=110−120=−10

( 1-1/2) : (1-1/3) : ( 1-1/4) = 1/2 : 2/3 : 3/4
= \(\dfrac{1}{2}\times\dfrac{3}{2}\times\dfrac{4}{3}\text{=}1\)
\(\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\)
\(\text{=}\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\text{=}\dfrac{5}{2}\)

học thôi
nếu banj ấy gỏi hết tất cả những môn khác thì nên kêu bạn ấy bớt học nhưng môn bạn đã giỏi và truyên tâm ôn đi ôn lại nhưng học kém

\(6\times\dfrac{3}{5}=\dfrac{18}{5};\dfrac{7}{2}+\dfrac{3}{10}=\dfrac{35}{10}+\dfrac{3}{10}=\dfrac{38}{10}=\dfrac{19}{5}\\ Ta.có:\dfrac{1}{10}< \dfrac{1}{2}< \dfrac{3}{5}< \dfrac{7}{10}< \dfrac{9}{10}< \dfrac{9}{7}< \dfrac{9}{5}< 3< \dfrac{18}{5}< \dfrac{19}{5}\\ \Leftrightarrow\dfrac{1}{10}< \dfrac{1}{2}< \dfrac{3}{5}< \dfrac{7}{10}< \dfrac{9}{10}< \dfrac{9}{7}< \dfrac{9}{5}< 3< 6\times\dfrac{3}{5}< \dfrac{7}{2}+\dfrac{3}{10}\)
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 :
\(\sqrt{2x-2}< \sqrt{2x-1}\left(x\ge2\right)\)
\(\Rightarrow A< 1\left(đpcm\right)\)