Bài 2:

a: \(\sqrt{\dfrac{130}{200}}:\dfrac{\sqrt{5}}{\sqrt{8}}=\dfrac{\sqrt{130}}{10\sqrt{2}}\cdot\dfrac{2\sqrt{2}}{\sqrt{5}}=\dfrac{\sqrt{26}}{5}\)

b: \(=\sqrt{\dfrac{1}{5}:\dfrac{4}{5}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

c: \(=\left(3\sqrt{5}+3\sqrt{5}-5\sqrt{5}\right):3\sqrt{5}\)

\(=\dfrac{\sqrt{5}}{3\sqrt{5}}=\dfrac{1}{3}\)

Sửa đề: Đưa thừa số vào trong dấu căn

a: \(3\sqrt{x^2}=\sqrt{3^2\cdot x^2}=\sqrt{9x^2}\)

b: \(-5\sqrt{y^4}=-\sqrt{5^2\cdot y^4}=-\sqrt{25y^4}\)

c: \(3\sqrt{5x}=\sqrt{3^2\cdot5x}=\sqrt{45x}\)

d: \(x\sqrt{7}=\sqrt{x^2\cdot7}=\sqrt{7x^2}\)

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: 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: \(A=\sqrt{16}+\sqrt{81}=4+9=13\)

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

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

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

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

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

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

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

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

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

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

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

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

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

a: \(3\sqrt{200}=3\cdot10\sqrt{2}=30\sqrt{2}\)

b: \(-5\sqrt{50a^2b^2}=-5\cdot5\sqrt{2a^2b^2}\)

\(=-25\cdot\left|ab\right|\cdot\sqrt{5}\)

c: \(-\sqrt{75a^2b^3}\)

\(=-\sqrt{25a^2b^2\cdot3b}=-5\left|ab\right|\cdot\sqrt{3b}\)

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)