a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)

b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)

c)Tương tự câu b, ta đc:

\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)

d)Tương tự câu a, ta đc:

\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)

Chúc Bạn Học Tốt!!!

a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)

b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)

c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)

d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)

\(\dfrac{\sqrt{3^2+39^2}}{\sqrt{7^2}+\sqrt{91^2}}\)

\(=\dfrac{1524}{7+91}=\dfrac{1524}{98}=\dfrac{762}{49}\)

=\(\dfrac{3\sqrt{170}}{7+9}\)

=\(\dfrac{1\cdot\sqrt{100+70}}{7+3}\)

=

\(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{7^2}+\sqrt{91^2}}\)\(=\frac{3+39}{7+91}=\frac{42}{98}=\frac{3}{7}\)

\(\dfrac{\sqrt{3^2}+39^2}{\sqrt{7^2+91^2}}\)=\(\dfrac{\sqrt{42^2}}{\sqrt{98^2}}=\dfrac{42}{98}=\dfrac{3}{7}\)

a)\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}.1\)

=\(\frac{3+39}{7+91}\)

=\(\frac{42}{98}\)

=\(\frac{3}{7}\)

b)\(\sqrt{\left(2,5-0,7\right)^2}\)

=\(|2,5-0,7|\)

=2,5-0,7

=1,8

a: \(=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2}\cdot\dfrac{1}{4}=\dfrac{1}{100}\)

b: \(=\dfrac{\left[5^3\left(5-1\right)\right]^3}{5^{12}}=\dfrac{5^9}{5^{12}}\cdot\dfrac{4^3}{1}=\dfrac{4^3}{5^3}\)

c: \(=\sqrt{1.8^2}=1.8\)

\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)

#Giải:

a)\(\sqrt{27}\)+\(\sqrt{75}\)-\(\sqrt{\dfrac{1}{3}}\)=8\(\sqrt{3}\)-\(\sqrt{\dfrac{1}{3}}\)=\(\dfrac{23\sqrt{3}}{3}\).

b)\(\sqrt{4+2\sqrt{3}}\)-\(\sqrt{4-2\sqrt{3}}\)=2.

c)\(\dfrac{3}{\sqrt{7}+\sqrt{2}}\)+\(\dfrac{2}{3+\sqrt{7}}\)+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=1,093+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=2,507.

a) = \(3\sqrt{3}+5\sqrt{3}-\dfrac{1}{\sqrt{3}}\)

= \(3\sqrt{3}+5\sqrt{3}-\dfrac{3}{\sqrt{3}}\)

= \(\dfrac{23\sqrt{3}}{3}\)

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

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

= \(1+\sqrt{3}-\sqrt{3}+1\)

= 2

c) = \(\dfrac{3\left(\sqrt{7}-\sqrt{2}\right)}{5}+\dfrac{2\left(3-\sqrt{7}\right)}{2}+\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)\)

= \(3\sqrt{7}-3\sqrt{2}+3-\sqrt{7}+2\sqrt{2}+2-2-\sqrt{2}\)

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

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

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

\(\approx2,5\)