a)\(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)

b) \(\sqrt{75.48}=\sqrt{25.3.16.3}=\sqrt{5^2.3^2.4^2}=5.4.3=60\)

c)\(\sqrt{90.6,4}=\sqrt{10.9.4.1,6}=\sqrt{4^2.3^2.2^2}=4.3.2=24\)

d) \(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}}=\sqrt{\left(\dfrac{5.12}{10}\right)^2}=\dfrac{5.12}{10}=6\)

a) \(\sqrt{45.80}=\sqrt{9.400}=\sqrt{9}.\sqrt{400}=3.20=60\)

b)\(\sqrt{75.48}=\sqrt{25.3.3.16}=5.3.4=60\)

c)\(\sqrt{90.6,4}=\sqrt{9.64}=3.8=24\)

d)\(\sqrt{2,5.14,4}=\sqrt{\dfrac{25}{10}.\dfrac{144}{10}}=\sqrt{\dfrac{25.144}{100}=\dfrac{5.12}{10}=\dfrac{60}{10}=6}\)

\(\sqrt{2 , 5.14 , 4}=\sqrt{25\cdot1,44}=\sqrt{25}\cdot\sqrt{1,44}=5\cdot1,2=6\)

d, \(\sqrt{2,5.14,4}=6\)

chúc bn học tốt

\(\sqrt{2,5.14,4}=\sqrt{36}=6\)

a) \(\sqrt{10}.\sqrt{40}\)

=\(\sqrt{10.40}\)

=\(\sqrt{400}\)

=20

b) \(\sqrt{5.}\sqrt{45}\)

=\(\sqrt{5.45}\)

=\(\sqrt{225}\)

=\(\sqrt{15}\)

c) \(\sqrt{52.}\sqrt{13}\)

=\(\sqrt{52.13}\)

=\(\sqrt{676}\)

=26

d)\(\sqrt{2.}\sqrt{162}\)

=\(\sqrt{2.162}\)

=\(\sqrt{324}\)

=18

b)

=15

a) \(\sqrt{10}.\sqrt{40}=\sqrt{10.40}=\sqrt{400}=20\)

b) \(\sqrt{5}.\sqrt{45}=\sqrt{5.45}=\sqrt{225}=15\)

c) \(\sqrt{52}.\sqrt{13}=\sqrt{52.13}=\sqrt{676}=26\)

d) \(\sqrt{2}.\sqrt{162}=\sqrt{2.162}=\sqrt{324}=18\)

@Akai Haruma

help em với

Nhầm môn rồi bạn ơi!!!

a.

$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$

$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$

$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$

$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$

$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$

---------------------

$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$

$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$

$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$

$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$

\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)

Ta có :

 \(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)

__

C làm giống câu a, nhé.

__

\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)

__

\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)

c.

$C=\sqrt{(2\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-2)^2}$

$=|2\sqrt{5}+1|-|\sqrt{5}-2|=2\sqrt{5}+1-(\sqrt{5}-2)=\sqrt{5}+3$

d.

$D=\sqrt{52-16\sqrt{3}}+\sqrt{4\sqrt{3}-7)^2}$

$=\sqrt{(4\sqrt{3})^2-2.4\sqrt{3}.2+2^2}+|4\sqrt{3}-7|$

$=\sqrt{(4\sqrt{3}-2)^2}+|4\sqrt{3}-7|$

$=|4\sqrt{3}-2|+|4\sqrt{3}-7|$

$=4\sqrt{3}-2+7-4\sqrt{3}=5$

a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)

\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)

=7-2

=5

d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)

\(=4\sqrt{7}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{4^2-2\cdot4\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(A=\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}\)

\(A=\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)\)

\(A=4^2-3\)

\(A=13\)

\(B=\dfrac{3}{4+\sqrt{13}}+\dfrac{\sqrt{52}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{\left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right)}+\dfrac{2\sqrt{13}}{2}-3\)

\(B=\dfrac{3\left(4-\sqrt{13}\right)}{16-13}+\sqrt{13}-3\)

\(B=4-\sqrt{13}+\sqrt{13}-3\)

\(B=4-3\)

\(B=1\)