\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\dfrac{\sqrt{3}-1}{4}\right)^2}=\dfrac{\sqrt{12+2\sqrt{3}}}{4}\)

\(\Rightarrow2\cos\alpha=\dfrac{\sqrt{12+2\sqrt{3}}}{2}\). Chọn B.

\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)

\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)

\(=4\sqrt{2}\)

\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

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

\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
 

Lời giải:

a. 

\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)

\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)

b.

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

c.

\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)

`a)A=2sqrt2-3sqrt{18}+4sqrt{32}-sqrt{50}`

`=2sqrt2-3sqrt{9.2}+4sqrt{16.2}-sqrt{25.2}`

`=2sqrt2-9sqrt2+16sqrt2-5sqrt2`

`=4sqrt2`

`b)B=sqrt{(1-sqrt5)^2}+sqrt{6+2sqrt5}`

`=sqrt5-1+sqrt{(sqrt5+1)^2}`

`=sqrt5-1+sqrt5+1=2sqrt5`

`c)1/(2-sqrt6)+1/(2+sqrt6)`

`=(2+sqrt6)/(4-6)+(sqrt6-2)/(6-4)`

`=(sqrt6-2-sqrt6-2)/2=-2`

`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`

`= (5+3-1)sqrt(x^2+2)=7sqrt6`

`<=> 7sqrt(x^2+2)=7sqrt6`.

`<=> x^2+2=36`.

`<=> x^2=34`.

`<=> x=+-sqrt(34)`.

Vậy...

`b, sqrt(4x^2-12x+9)-6=0`

`<=> |2x-3|=6`.

`@ x >=3/2 <=> 2x-3=6.`

`<=> x=9/2 (tm)`.

`@x <3/2 <=> 3-2x=6`

`<=> 2x=-3`

`<=> x=-3/2.`

Vậy...

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)