a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)

\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)

\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)

\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)

\(=-8\sqrt{3}\)

b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)

\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)

Rút gọn rồi tính:                                                                                                                                               \(a.\frac{16}{24}-\frac{1}{3}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)

\(b.\frac{4}{5}-\frac{12}{60}=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}\)

Tính rồi rút gọn:

\(a.\frac{17}{6}-\frac{2}{6}=\frac{15}{6}\)

\(b.\frac{16}{15}-\frac{11}{15}=\frac{5}{15}=\frac{1}{3}\)

\(c.\frac{19}{12}-\frac{13}{12}=\frac{6}{12}=\frac{1}{2}\)

1a2/3-1/3=1/3            c4/5-1/5=3/5             tinh roi rut gon 17/6-2/6=15/6=5/2             16/15-11/15=5/15=1/3           19/12-13/12=6/12=1/2

chuc bn hoc tot

lớp 1 à cu

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

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

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

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

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

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

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

\(=2\sqrt{2}-2+2\sqrt{2}+1\)

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

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

sao bn ko tách ra

sao bn khum tách ra,làm cho 1 tăng thế ai chả lời

tách ra mệt

A) \(\dfrac{5}{10}-\dfrac{2}{15}\)

\(=\dfrac{1}{2}-\dfrac{2}{15}\)

\(=\dfrac{15}{30}-\dfrac{4}{30}\)

\(=\dfrac{11}{30}\)

B)\(\dfrac{5}{20}-\dfrac{1}{6}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}\)

\(=\dfrac{6}{24}-\dfrac{4}{24}\)

\(=\dfrac{2}{24}=\dfrac{1}{12}\)

C) \(\dfrac{6}{18}-\dfrac{6}{24}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}\)

\(=\dfrac{4}{12}-\dfrac{3}{12}\)

\(=\dfrac{1}{12}\)

D) \(\dfrac{5}{9}-\dfrac{3}{12}\)

\(=\dfrac{5}{9}-\dfrac{1}{4}\)

\(=\dfrac{20}{36}-\dfrac{9}{36}\)

\(=\dfrac{11}{36}\).

a: =1/2-2/15

=15/30-4/30

=11/30

b: =1/4-1/6

=3/12-2/12

=1/12

c: =1/3-1/4

=1/12

d: =20/36-9/36

=11/36

Rút gọn rồi tính:

a) $\frac{16}{24}–\frac{1}{3}=\frac{2}{3}–\frac{1}{3}=\frac{2–1}{3}=\frac{1}{3}$     

b) $\frac{4}{5}–\frac{12}{60}=\frac{4}{5}–\frac{1}{5}=\frac{3}{5}$

Tính rồi rút gọn:

a) $\frac{17}{6}–\frac{2}{6}=\frac{17–2}{6}=\frac{15}{6}=\frac{5}{2}$             

b) $\frac{16}{15}–\frac{11}{15}=\frac{16–11}{15}=\frac{5}{15}=\frac{1}{3}$

c)