\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)

a. (

a) Ta có A = 2 x + 1 x : 2 x − 1 x = 2 x + 1 2 x − 1  

b) Ta có  B = a + 2 a − 2 : a 2 + 4 a + 4 a 2 + 2 a + 4 = a + 2 a − 2 . a 2 + 2 a + 4 ( a + 2 ) 2 = a 2 + 2 a + 4 a 2 − 4

a)     m m − 2 − m m + 2 m + 2 m m − 2 m = m + 2 m − 2

b)     3 5 − 3 m + 1 16 − m 2 m 2 + 2 m + 1 = 3 m − 12 5 ( m − 1 ) 16 − m 2 ( m + 1 ) 2 = − 3 ( m + 1 ) 5 ( m + 4 )

Đặt \(A=\sqrt{2+\sqrt{3}}\)

    \(\Rightarrow A\sqrt{2}=\sqrt{4+2\sqrt{3}}\)

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

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

                        \(=\sqrt{3}+1\)

\(\Rightarrow A=\frac{\sqrt{3}+1}{2}hay\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{2}\)

TK nha!

\(\sqrt{2+\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}\)

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

                           \(=\frac{\sqrt{3}+1}{\sqrt{2}}\)

                              \(=\frac{\sqrt{6}+\sqrt{2}}{2}\)

Lời giải:
$\sqrt{0,01}-\sqrt{0,25}=0,1-0,5=-0,4$

$0,5\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-\frac{1}{2}=\frac{9}{2}$

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\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)

Giải:

1) \(\sqrt{25-4\sqrt{6}}\)

\(=\sqrt{24-4\sqrt{6}+1}\)

\(=\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.1+1^2}\)

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

\(=2\sqrt{6}-1\)

Vậy ...

1 \(\sqrt{25-4\sqrt{6}}=\sqrt{24-4\sqrt{6}+1}\)

= \(\sqrt{\left(\sqrt{24}-1\right)^2}=\sqrt{24-1}\)