1: \(\left(\sqrt{15}-2\sqrt3\right)^2+12\sqrt5\)

\(=15+12-2\cdot\sqrt{15}\cdot2\sqrt3+12\sqrt5\)

\(=27-4\sqrt{45}+12\sqrt5=27\)

2: \(3\sqrt2\left(4-\sqrt2\right)+3\left(1-2\sqrt2\right)^2\)

\(=12\sqrt2-6+3\left(9-4\sqrt2\right)\)

\(=12\sqrt2-6+27-12\sqrt2=21\)

3: \(\frac12\left(\sqrt6+\sqrt5\right)^2-\frac14\cdot\sqrt{120}-\sqrt{\frac{15}{2}}\)

\(=\frac12\left(11+2\sqrt{30}\right)-\frac14\cdot2\sqrt{30}-\sqrt{\frac{30}{4}}\)

\(=\frac{11}{2}+\sqrt{30}-\frac12\sqrt{30}-\frac12\sqrt{30}=\frac{11}{2}\)

4: \(\left(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7}\right)^2\)

\(=4-\sqrt7+4+\sqrt7-2\cdot\sqrt{\left(4-\sqrt7\right)\left(4+\sqrt7\right)}\)

\(=8-2\cdot\sqrt{16-7}=8-2\cdot\sqrt9=8-2\cdot3=8-6=2\)

5: \(\left(\sqrt{\sqrt{14}+\sqrt5}+\sqrt{\sqrt{14}-\sqrt5}\right)^2\)

\(=\sqrt{14}+\sqrt5+\sqrt{14}-\sqrt5+2\cdot\sqrt{\left(\sqrt{14}+\sqrt5\right)\left(\sqrt{14}-\sqrt5\right)}\)

\(=2\sqrt{14}+2\cdot\sqrt{14-5}=2\sqrt{14}+2\cdot\sqrt9=2\sqrt{14}+6\)

6: \(\left(\sqrt3+1\right)^3-\left(\sqrt3-1\right)^3\)

\(=\left(3\sqrt3+3\cdot3\cdot1+3\cdot\sqrt3\cdot1^2+1\right)-\left(3\sqrt3-3\cdot3\cdot1+3\sqrt3-1\right)\)

\(=\left(6\sqrt3+10\right)-\left(6\sqrt3-10\right)=20\)

7: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left\lbrack\left(\sqrt2\right)^3+3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2+1^3\right\rbrack-\left\lbrack\left(\sqrt2\right)^3-3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2-1^3\right\rbrack\)

\(=\left(2\sqrt2+6+3\sqrt2+1\right)-\left(2\sqrt2-6+3\sqrt2-1\right)=5\sqrt2+7-5\sqrt2+7=14\)

8: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+2\cdot\sqrt{360}}\)

\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}-\sqrt{45+2\cdot3\sqrt5\cdot2\sqrt2+8}\)

\(=\sqrt{\left(2\sqrt2-\sqrt5\right)^2}-\sqrt{\left(3\sqrt5+2\sqrt2\right)^2}\)

\(=2\sqrt2-\sqrt5-3\sqrt5-2\sqrt2=-4\sqrt5\)

9: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)\)

\(=\frac12\left(\sqrt5-1+\sqrt5+1\right)=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

1/ \(=2+\sqrt{5}-\left|2-\sqrt{5}\right|=2+\sqrt{5}-\sqrt{5}+2=4\)

2/ bạn coi lại đề

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

4/ \(=\sqrt{3}+2-\left|\sqrt{3}-2\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)

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

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

Các bạn giúp mình với, tối nay mình nộp rồi.

Câu 6 sửa lại đề giúp mình như này nhé:

\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

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

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

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

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

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

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

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

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

b, \(2\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}+1\right)^2}\)

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

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

c, \(4\sqrt{\left(2-\dfrac{\sqrt{3}}{2}\right)^2}-3\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=4\left|2-\dfrac{\sqrt{3}}{2}\right|-3\left|\sqrt{3}-1\right|\)

\(=8-2\sqrt{3}-3\sqrt{3}+3=11-5\sqrt{3}\)

c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)

\(=4+\sqrt{10}-4+\sqrt{10}\)

\(=2\sqrt{10}\)

d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)

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

\(=2\sqrt{2}\)

a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)

b) \(=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}-\sqrt{2015}=-\sqrt{2013}-\sqrt{2015}\)

c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)

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

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

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

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)

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