\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}=\sqrt{20\cdot72\cdot4,9}=\sqrt{2\cdot10\cdot72\cdot4,9}\\ =\sqrt{144\cdot49}=\sqrt{144}\cdot\sqrt{49}=12\cdot7=84\)

Bài 2:

a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

√(2a . 32ab2) = √(64a2b2 )

= √((8ab)2) = 8ab (do a ≥ 0; b ≥ 0)

a )   √ ( 3 a 3   ) . √ 12 a   =   √ ( 3 a 3 . 12 a )   =   √ ( 36 a 4   )     =   √ ( ( 6 a ²   ) 2   )   =   6 a 2   ( d o   a 2   ≥   0 )     b )   √ ( 2 a   .   32 a b 2 )   =   √ ( 64 2 b 2   )     =   √ ( ( 8 a b ) 2 )   =   8 a b   ( d o   a   ≥   0 ;   b   ≥   0 )

a) M = 8ab;

b) N = [ ( 3 a   + +   2 )   +   ( 1   –   2 b ) ] 2   =   ( 3 a   –   2 b   +   3 ) 2 .

√(3a3 ).√12a = √(3a3.12a) = √(36a4 )

= √((6a2 )2 ) = 6a2 (do a2 ≥ 0)

\(A=\frac{x+3a}{2-x}+\frac{x-3a}{2+x}-\frac{2a}{4-x^2}+a\)

\(=\frac{\left(x+3a\right)\left(2+x\right)+\left(x-3a\right)\left(2-x\right)-2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\frac{2x+x^2+6a+3ax+2x-x^2-6a+3ax-2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\frac{6ax+4x-2a+a\left(4-x^2\right)}{\left(2-x\right)\left(2+x\right)}=\frac{6ax+4x-2a+4a-a\cdot x^2}{\left(2-x\right)\left(2+x\right)}=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(-a\cdot x^2+6a\cdot x+2a\)

\(=-a\cdot\frac{a^2}{\left(3a+2\right)^2}+6a\cdot\frac{a}{3a+2}+2a=\frac{-a^3}{\left(3a+2\right)^2}+\frac{6a^2\left(3a+2\right)+2a\left(3a+2\right)^2}{\left(3a+2\right)^2}\)

\(=\frac{-a^3+18a^3+12a^2+2a\left(9a^2+12a+4\right)}{\left(3a+2\right)^2}\)

\(=\frac{17a^3+12a^2+18a^3+24a^2+8a}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}\)

\(\left(2-x\right)\left(2+x\right)\)

\(=\left(2-\frac{a}{3a+2}\right)\left(2+\frac{a}{3a+2}\right)=\frac{6a+4-a}{\left(3a+2\right)}\cdot\frac{6a+4+a}{3a+2}\)

\(=\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}\)

\(A=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}:\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(5a+4\right)\left(7a+4\right)}\)

Ta có : 

\(\sqrt{2a.32ab^2}\)

\(=\)\(\sqrt{64a^2b^2}\)

\(=\)\(\sqrt{8^2a^2b^2}\)

\(=\)\(\sqrt{\left(8ab\right)^2}\)

\(=\)\(\left|8ab\right|\)

Chúc bạn học tốt ~ 

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

b)\(\sqrt{\left(2-\sqrt{11}\right)^2}=2-\sqrt{11}\)

c)\(2\sqrt{a^2}=2a\) vì a≥0