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

Các số bằng \(\dfrac{3}{7}\) là a ; b ; c ; d

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

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

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

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

Giải:

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

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

\(=\left|1,8\right|=1,8\)

Vậy ...

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

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

\(=\dfrac{42}{98}=\dfrac{3}{7}\)

Vậy ...

\(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\frac{3-39}{7-91}=\frac{-36}{-84}=\frac{3}{7}\)

Happy memories -_- là \(\sqrt{7^2}-\sqrt{91^2}\)thành \(\sqrt{7^2-\sqrt{91^2}}\) chứ

3/592

k nha

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

0100101010100101010010101010010010101001010100101