1: =3+căn 2-3+căn 2

=2căn 2

2: =(căn 3-2)(căn 3+2)

=3-4=-1

Sửa đề: \(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)

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

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

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

=3-1=2

Hihi mình cũng học lớp 9, để mình giúp cậu nha!

a) \(\sqrt{3-2\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(1+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{2}-1\right|+\left|1+\sqrt{8}\right|=\sqrt{2}-1+1+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

b) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}\)

\(=\left|\sqrt{7}-1\right|-\left|1+\sqrt{7}\right|=\sqrt{7}-1-1-\sqrt{7}=-2\)

c) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

(Nhớ click cho mình với nhoa!)

A=3³=27
\(B=\dfrac{2^{11}.9^2}{3^5.2^8}=\dfrac{2^{11}.3^4}{3^5.2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)

A=3^3=9

B=2^3/3^5=8/243

\(\sqrt{11+6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)

Mik ko hiểu đề lắm

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}\right)\)

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

\(=\sqrt{6}-3\)