a) \(\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}.\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\)

b) \(\sqrt{\frac{3a}{5b}}=\frac{\sqrt{3a}}{\sqrt{5b}}=\frac{\sqrt{3a}.\sqrt{5b}}{5b}=\frac{\sqrt{15ab}}{5b}\left(a;b>0\right)\)

c) \(\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{12}}=\frac{\sqrt{5}.\sqrt{12}}{12}=\frac{\sqrt{60}}{12}=\frac{2\sqrt{15}}{12}=\frac{\sqrt{15}}{6}\)

d) \(\sqrt{\frac{5x}{18y}}=\frac{\sqrt{5x}}{\sqrt{18y}}=\frac{\sqrt{5x}}{\sqrt{3^2.2y}}=\frac{\sqrt{5x}}{3\sqrt{2y}}\)

\(=\frac{\sqrt{5x}.\sqrt{3y}}{3.2y}=\frac{\sqrt{15xy}}{6xy}\)

Quên mất k ghi đk xy > 0

a) 2√6>3√2>√13>2√326

b)1/3√39>1/4√32>1/5√35>1/2√51339

@@@

Bạn Tạ Bảo Trân làm sai

a) Ta sắp xếp theo thứ tự tăng dần như sau:

\(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)

b) Ta sắp xếp theo thứ tự tăng dần như sau:

\(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)

Hoàng Phong làm bừa

đề bài là gì ạ

so sánh hay gì ạ

....

a) Ta có:

\(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}.\)

Mà \(\sqrt{12}< \sqrt{13}\)

Nên \(2\sqrt{3}< \sqrt{13}\) 

a, \(-\frac{2}{3}\sqrt{ab}=-\sqrt{\frac{4ab}{9}}\)

b, \(a\sqrt{\frac{3}{a}}=\sqrt{\frac{3a^2}{a}}=\sqrt{3a}\)

c, \(a\sqrt{7}=\sqrt{7a^2}\)

d, \(b\sqrt{3}=\sqrt{3b^2}\)

e, \(ab\sqrt{\frac{a}{b}}=\sqrt{\frac{a^3b^2}{b}}=\sqrt{a^3b}\)

f, \(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a^2b^2}{a}+\frac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)

a, −23√ab=−√4ab9−23ab=−4ab9

b, a√3a=√3a2a=√3aa3a=3a2a=3a

c, a√7=√7a2a7=7a2

d, b√3=√3b2b3=3b2

e, ab√ab=√a3b2b=√a3babab=a3b2b=a3b

f, ab√1a+1b=√a2b2a+a2b2b=√ab2+a2b

a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)

\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)

\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)

\(=\left(4-8+25\right)\sqrt{x^2+1}\)

\(=21\sqrt{x^2+1}\)

b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)

\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)

\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)

\(B=\sqrt{3}\)

a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}-18\sqrt{y+4}=-7\sqrt{y+4}\)

c) \(P=\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}=5\sqrt{y-2}\)

a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}.\)

\(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}\)

\(=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}\)

\(=-7\sqrt{y+4}\)

c) \(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}\)

\(=5\sqrt{y-2}\)

a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)

\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)

\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)

\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)

\(A=\left(4-12+5\right)\sqrt{2}\)

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

b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)

\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)

\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)

\(B=2\sqrt{3}\)

c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)

\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)

\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)

\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)

\(C=\left(2+12-10\right)\sqrt{5a}\)

\(C=4\sqrt{5a}\)

a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\)                               \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\)                                                                                              b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)