\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)

\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)

\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)

a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)

b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)

c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

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

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)

a) \(\sqrt{27x^2}\)

\(=\sqrt{3^2\cdot3x^2}\)

\(=\left|3x\right|\sqrt{3}\)

\(=3\left|x\right|\sqrt{3}\)

b) \(\sqrt{8xy^2}\)

\(=\sqrt{2^2\cdot2\cdot x\cdot y^2}\)

\(=\left|2y\right|\sqrt{2x}\)

\(=2\left|y\right|\sqrt{2x}\)

c) \(\sqrt{25x^3}\)

\(=\sqrt{5^2\cdot x^2\cdot x}\)

\(=\left|5x\right|\sqrt{x}\)

\(=5\left|x\right|\sqrt{x}\)

d) \(\sqrt{48xy^4}\)

\(=\sqrt{4^2\cdot3x\cdot\left(y^2\right)^2}\)

\(=\left|4y^2\right|\sqrt{3x}\)

\(=4y^2\sqrt{3x}\)

`a, sqrt(27x^2b) = sqrt(3^2. 3.x^2b) = 3|x|sqrt(3b)`.

`b, sqrt(8xy^2) =sqrt(2^2.2xy^2)= 2|y|sqrt(2x)`

`c, sqrt(25x^3d) = sqrt(5^2.x^2.x.d) = 5|x|sqrt(xd)`.

`d, sqrt(48xy^4) = sqrt(4^2.3 . xy^4) = 4y^2sqrt(3x)`.

a) \(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}=2\sqrt{196}-10\sqrt{245}+14\sqrt{784}\)

\(=28-10\sqrt{49.5}+392=420-70\sqrt{5}\)

b) \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}=\sqrt{144}-3\sqrt{48}+5\sqrt{16}+4\sqrt{9.3}\)

\(=12-3\sqrt{16.3}+20+12\sqrt{3}=32-12\sqrt{3}+12\sqrt{3}=32\)

\(\frac{1}{3}\sqrt{9+6a+a^2}+\frac{4a}{3}+5\)

\(=\frac{1}{3}\sqrt{\left(a+3\right)^2}+\frac{4a}{3}+5\)

\(=\frac{1}{3}\left|a+3\right|+\frac{4a}{3}+5\)(1)

Với a < 3 \(\left(1\right)=-\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=a+4\)

Với a >= 3 \(\left(1\right)=\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=\frac{5}{3}a+6\)

đéo biết làm, đăng làm quái gì không biết -_-"

a) \(\sqrt{\frac{9a^2-12ab+4b^2}{81a^4b^4}}=\sqrt{\frac{\left(3a-4b\right)^2}{\left(9a^2b^2\right)^2}}\)

\(=\frac{3a-4b}{9a^2b^2}\)

b)\(\sqrt{\frac{1}{a}-\frac{1}{a^2}}=\sqrt{\frac{a-1}{a^2}}=\frac{1}{a}\sqrt{a-1}\)

P/s tham khảo nhé