Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)

\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)

Vậy A<B

Hầu hết các dạng bài này bạn chỉ cần quy đồng là ra ngay nhé :)

Điều kiện xác định : \(0< x\ne1\)

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

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

giúp tớ với

Ta có : \(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

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

\(M=\frac{\sqrt{a}-1}{\sqrt{a}}\)

\(M=1-\frac{1}{\sqrt{a}}\)

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

Điều kiện x \(\ge\frac{1}{4}\)

Đặt a = \(\sqrt{x-\frac{1}{4}}\)(a \(\ge0\))

=> x = a² + \(\frac{1}{4}\)

=> PT <=> 2a² + \(\frac{1}{2}\)+ \(\sqrt{a^2+\frac{1}{4}+a}\)= 2

<=> \(\sqrt{a^2+\frac{1}{4}+a}\)= \(\frac{3}{2}-2a\)

<=> a² + 0,25 + a = 4a⁴ + 2,25 - 6a²

<=> 4a⁴ - 7a² - a + 2 = 0

<=> (a + 1)(2a - 1)(2a² - a - 2) = 0

<=> a = 0,5

<=> x = 0,5

a) 

= \(\sqrt{18-6\sqrt{6}+3}\)        

= \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)       

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

= \(|3\sqrt{2}-\sqrt{3}|\)   

= \(3\sqrt{2}-\sqrt{3}\)   

b) 

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

= \(\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)    

= \(\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)     

= \(|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}|\) 

= \(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\)        

c) 

= \(\sqrt{3+2\sqrt{3}+1}\)  

= \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}\)    

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

d) 

Đặt t = \(\sqrt{x-1}\left(ĐK:t\ge0\right)\)   

= \(\sqrt{t^2+1-2t}\)       

= \(\sqrt{\left(t+1\right)^2}\)   

\(=t+1\)      

= \(\sqrt{x-1}+1\)                     

\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2\sqrt{9}\sqrt{6}+3}=\sqrt{\left(\sqrt{18}\right)^2-2\sqrt{18}\sqrt{3}+\left(\sqrt{3}\right)^2}\)

                                \(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}=\sqrt{3}+3\sqrt{2}\)

\(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)

                           \(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{2}}{2}\)

\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Với \(x\ge1\)thì \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)

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

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

T đã tốn mấy phút cuộc đời viết lời giải cho bạn r, tiếc j mấy giây mà bấm k cho t ik =))

\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)

....... Tới đây được chưa bạn?