\(\sqrt{96}.\sqrt{125}\)

\(\sqrt{16.6}\sqrt{25.5}\)

\(4.5\sqrt{6.5}\)

\(20\sqrt{30}\)

\(b,\sqrt{a^4b^5}\)

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

\(c,\sqrt{a^6b^{11}}\)

\(a^3b^5\sqrt{b}\)

\(d,\sqrt{a^3\left(1-a\right)^4}\)

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

 a/   \(\sqrt{a^4b^5}=a^2b^2\sqrt{b}\)

b/    \(\sqrt{a^6b^{11}}=a^3b^5\sqrt{b}\)

\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)

\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)

\(\sqrt{225\cdot17}=15\sqrt{17}\)

=\(\sqrt{16\cdot6\cdot25\cdot5}\)

=\(\sqrt{4^2\cdot6\cdot5^2\cdot5}\)

=4*5\(\sqrt{6\cdot5}\)

=20\(\sqrt{30}\)

b) =\(\sqrt{\left(a^2\right)^2\cdot\left(b^2\right)^2\cdot b}\)

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

d)=\(\sqrt{11\cdot11\cdot4\cdot a^2}\)

=\(\sqrt{11^2\cdot2^2\cdot a^2}\)

=11*2*a

=22a

a/ \(\sqrt{50a}=5\sqrt{2a}\)

b/ \(\sqrt{75x}=5\sqrt{3x}\)

a)\(5\sqrt{2a}\)

b)\(5\sqrt{3x}\)

a) \(\sqrt{50a}=\sqrt{25.2}=5\sqrt{2a}\)

b) \(\sqrt{75x}=\sqrt{25.3x}=5\sqrt{3x}\)

\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)

\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)

\(=3\sqrt{2}\left(2a-1\right)\cdot b\sqrt{b}\)

đưa thừa số vào trong dấu căn*

mình nhầm

a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)

b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)

\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)

a: ĐKXĐ: 2x-10>=0

=>2x>=10

=>x>=5

b: \(\sqrt{A^2B}=\sqrt{A^2}\cdot\sqrt{B}=\left|A\right|\cdot\sqrt{B}\)

\(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}\)

c: \(A=\sqrt{16}+\sqrt{81}=4+9=13\)

\(B=\sqrt{\dfrac{\left(15\sqrt{5}+5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}}\)

\(=\sqrt{\dfrac{15}{\sqrt{2}}+5\sqrt{20}-3\sqrt{45}}\)

\(=\sqrt{\dfrac{15\sqrt{2}+2\sqrt{5}}{2}}=\sqrt{\dfrac{30\sqrt{2}+4\sqrt{5}}{4}}\)

\(=\dfrac{\sqrt{30\sqrt{2}+4\sqrt{5}}}{2}\)

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

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

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

chịu.-.

HT~~~

a, -2x^2y căn 7 

b, ab^2 căn 63 

c, a-1 căn 147a-147

d, y+2 nhân căn [192 nhân (y+2)^3]

a)-2x²y√7

b) 3ab²√7

c) 7(a-1)√3(a-1)

d) 8(y+2)²√3(y+2)