Question

triển khai các tích sau

a)( 1/2m+3)3

b)[2(m+1/2)]

c)(2 căn x+1)2 vớix>0

d)(2*căn bậc hai(2)+1/4)2giúp mình v

Answer 1

a) \(\left(\dfrac{1}{2}m+3\right)^3=\dfrac{m^3}{8}+\dfrac{9m^2}{4}+\dfrac{27m}{2}+9\)

b) \(2\left(m+\dfrac{1}{2}\right)=2m+1\)

c) \(\left(2\sqrt{x}+1\right)^2=4x+4\sqrt{x}+1\)

d) \(\left(2\sqrt{2}+\dfrac{1}{4}\right)^2=8+\sqrt{2}+\dfrac{1}{16}=\dfrac{129}{16}+\sqrt{2}\)

Answer 2

chi tiết câu c

c) \(\left(2\sqrt{x}+1\right)^2=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2=4x+4\sqrt{x}+1\)

Answer 3

a) \(\left(\dfrac{1}{2}m+3\right)^3\)

\(=\left(\dfrac{1}{2}m\right)^3+3\cdot\left(\dfrac{1}{2}m\right)^2\cdot3+3\cdot\dfrac{1}{2}m\cdot3^2+3^3\)

\(=\dfrac{1}{8}m^3+\dfrac{9}{4}m^2+\dfrac{27}{2}m+27\)

c) Ta có: \(\left(2\sqrt{x}+1\right)^2\)

\(=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2\)

\(=4x+4\sqrt{x}+1\)