a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)

b) \(-\sqrt{500.162}=-\sqrt{\left(2^2.5^2.5\right).\left(2.3^2.3^2\right)}\)

\(=-2.5.3.3\sqrt{5.2}=-90\sqrt{10}\)

c) \(\sqrt{125a^2}=\sqrt{5^2.a^2.5}=5\left|a\right|\sqrt{5}=-5a\sqrt{5}\) (do a <0)

d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|\)

a < 0 thì suy ra \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|=-5a\)

a>=0 thì suy ra \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{15}{3}\left|a\right|=5\left|a\right|=5a\)

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)

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

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

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

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

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

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

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

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

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

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 \(a\ge0;a\ne1\)

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

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

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

Bài 1: Đưa thừa số ra ngoài dấu căn:

\(2\sqrt{225a^2}=2.15a=30a\)

Bài 2: Đưa thừa số vào trong dấu căn :

\(x\sqrt{\dfrac{-39}{x}}=\sqrt{x^2.\dfrac{-39}{x}}=\sqrt{-39x}\)

Bài 3: Sắp xếp theo thứ tự tăng dần :

a) \(2\sqrt{3}< 3\sqrt{2}< 2\sqrt{5}< 5\sqrt{2}\)

b) \(4\sqrt{2}< \sqrt{37}< 2\sqrt{15}< 3\sqrt{7}\)

c) \(6\sqrt{\dfrac{1}{3}}< \sqrt{27}< 2\sqrt{28}< 5\sqrt{7}\)

\(1,ĐKXĐ:x\ge0\\ x\sqrt{3}=-\sqrt{3x^2}\\ \Leftrightarrow3x^2=9x^2\\ \Leftrightarrow6x^2=0\\ \Leftrightarrow x=0\left(tm\right)\)

\(2,ab^2\sqrt{a}=ab^2\sqrt{a}\)

\(3,a\sqrt{\dfrac{b}{a}}=\sqrt{ab}\)

\(ab^2\sqrt{a}=\sqrt{a^3b^4}\)

Bài 2: 

a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)

b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)