a) Ta có:

      \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)

b) Ta có:

\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

Bạn tham khảo câu số 9:

mọi người giúp em mấy bài này với ạ =((( - Hoc24

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

\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)

b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)

Vậy ta có đpcm 

b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}\)

\(=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1\)

b) Ta có: \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+\sqrt{a}+a\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}\right)^2:\left(1+\sqrt{a}\right)^2=1\)

a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)

=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)

=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)

=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))

=2(\(\sqrt{5}+5-\sqrt{5}-1\))

=2.4=8=VP
=> đpcm

b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)

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

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

=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)

vậy VT=VP =>đpcm

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

\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

\(=9=VP\)

2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

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

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

Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

\(\Rightarrow A< \frac{1}{2}\)

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

\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)

\(\Rightarrow A< 2\)

Bài 2 tạm thời chưa nghĩ ra :))