a, \(\frac{a}{\sqrt{a}}=\sqrt{a}\)

b, \(\frac{a}{\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)

c, \(\frac{x}{\sqrt{3x^3}}=\frac{x}{x\sqrt{3x}}=\frac{1}{\sqrt{3x}}=\frac{\sqrt{3x}}{3x}\)

d, \(\frac{4y^2}{\sqrt{2y^5}}=\frac{4y^2}{y^2\sqrt{2y}}=\frac{4}{\sqrt{2y}}=\frac{4\sqrt{2y}}{2y}=\frac{2\sqrt{2y}}{y}\)

a)\(\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}\)                b) \(\dfrac{a}{\sqrt{ab}}=\dfrac{a\sqrt{ab}}{\left(\sqrt{ab}\right)^2}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b}\)                                              c) \(\dfrac{x}{\sqrt{3x^3}}=\dfrac{x\sqrt{3x}}{\sqrt{3x^3.\sqrt{3x}}}=\dfrac{x\sqrt{3x}}{\left(\sqrt{3x^2}\right)^2}=\dfrac{x\sqrt{3x}}{\left(3x^2\right)^2}=\dfrac{x\sqrt{3x}}{3x^2}=\dfrac{\sqrt{3x}}{3x}\)    

a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\)  =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\)  -  \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+  \(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}.21a}\) -  \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\)  -   \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+  \(\sqrt{21a}\)

=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\)  +  \(\sqrt{21a}\)

=\(\frac{-10}{21}\sqrt{21a}\)

b)

N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)

=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)

=\(\frac{2}{3}\sqrt{6x}-\frac{3}{2}.\sqrt{6x}+\sqrt{6x}\)

=\(\frac{1}{6}\sqrt{6x}\)

em lớp 8 nene làm theo cách hiểu thôi ạ

Answer:

\((ac+bd)^2+(ad-bc)^2\)

\(=(ac)^2+(2acbd)+(bd)^2+(ad)^2-(2acbd)+(bc)^2\)

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

\(=a^2(c^2+d^2)+b^2(c^2+d^2)\)

\(=(c^2+d^2)(a^2+b^2)\)

tự làm

) $xy\sqrt{\genfrac{}{}{0ex}{}{x}{y}}=xy\sqrt{\genfrac{}{}{0ex}{}{xy}{y^2}}=\genfrac{}{}{0ex}{}{xy}{y}\sqrt{xy}=x\sqrt{xy}.$

b) $\genfrac{}{}{0ex}{}{x}{y}\sqrt{\genfrac{}{}{0ex}{}{x}{y}}=\genfrac{}{}{0ex}{}{x}{y}\sqrt{\genfrac{}{}{0ex}{}{xy}{y^2}}=\genfrac{}{}{0ex}{}{x}{y^2}\sqrt{xy}.$

c) $\sqrt{\genfrac{}{}{0ex}{}{1}{a}+\genfrac{}{}{0ex}{}{1}{a^2}}=\sqrt{\genfrac{}{}{0ex}{}{a+1}{a^2}}=\genfrac{}{}{0ex}{}{\sqrt{a+1}}{a}.$

d) $\sqrt{\genfrac{}{}{0ex}{}{4x^3}{25y}}=\sqrt{\genfrac{}{}{0ex}{}{4x^{2}xy}{25y^2}}=\genfrac{}{}{0ex}{}{2x}{5y}\sqrt{xy}.$

e) $2ab\sqrt{\genfrac{}{}{0ex}{}{3}{ab}}=2\sqrt{\genfrac{}{}{0ex}{}{3(ab{)}^2}{ab}}=2\sqrt{3ab}$.

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