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

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

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

\(\dfrac{11\cdot9^{11}\cdot3^7-27^{10}}{4\cdot81^1}\)

\(=\dfrac{11\cdot3^{22}\cdot3^7-\left(3^3\right)^{10}}{4\cdot3^4}\)

\(=\dfrac{11\cdot3^{29}-3^{30}}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot\left(11-3\right)}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot8}{4\cdot3^4}\)

\(=3^{25}\cdot2\)

@ Huỳnh Thanh Phong

Sao lại lấy 3²⁹ là thừa số chung ạ

đề bài là thế này đúng ko 11 .9¹¹ .3⁷ - 9⁵ : ( 2 . 3¹⁴)²

ta có : 11 . ( 3²)¹¹ . 3⁷- (3²)¹⁵ : 22 . 3²⁸

         = 11 . 3²² . 3⁷ - 3³⁰ : 2² .3²⁸ 

            = 11 . 3²⁹ - 3³⁰ : 2² . 3²⁸

            = 3²⁹. ( 11 - 3 ) : 2²  . 3²⁸

            = 3²⁹. 8 : 2² . 3²⁸

           = 3^{29 .} 2³ : 2^2 . 3²⁸

           = ( 3²⁹ : 3²⁸ ) . ( 2³ : 2² )

        = 3 . 2

       = 6

x=1;

x=1;

x=1.

k cho mình nhé.

dễ

\(\frac{11.9^{11}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

= \(\frac{11.\left(3^2\right)^{11}.3^7-\left(3^2\right)^{15}}{2^2.3^{28}}\)

= \(\frac{11.3^{22}.3^7-3^{30}}{4.3^{28}}\)

= \(\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)

= \(\frac{\left(11-3\right).3^{29}}{4.3^{28}}\)

= \(\frac{8.3^{29}}{4.3^{28}}\)

= 2 . 3

= 6

a) \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1=a^3+a^2b-a^2b+b^3+1=a^3+b^3+1\)

b) \(P=x\left(x-y+1\right)-y\left(y+1-x\right)-2=x^2-xy+x-y^2-y+xy-2=x^2+x-y-y^2-2\)

c) \(Q=\left(m+3\right)\left(m^2+3m-5\right)+\left(6-m\right)m^2+11=m^3+3m^2-5m+3m^2+9m-15+6m^2-m^3+11=12m^2+4m-4\)

a: Ta có: \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1\)

\(=a^3+a^2b-a^2b+b^3+1\)

\(=a^3+b^3+1\)

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

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

\(=\dfrac{4-2\sqrt{5}}{-1}+2\sqrt{5}\)

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

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

Ta có: \(\dfrac{2}{\sqrt{5}+2}+2\sqrt{5}\)

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

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

nhầm đề chỗ nào ko

phải có 1 dấu - chứ

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=8x^3+y^3\)