Ta có: \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{21}{35}+\dfrac{6}{30}\)

\(=\dfrac{-1}{3}+\dfrac{-2}{3}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{6}{30}\)

\(=-1+1+\dfrac{1}{5}\)

\(=\dfrac{1}{5}\)

a) A= \(\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}\)

= \(\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)\)

= \(\dfrac{4+2+1}{7}+\dfrac{3+5}{4}\)

= \(\dfrac{7}{7}+\dfrac{8}{4}\) = \(1+2\) = \(3\)

b) B= \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\)

= \(\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)

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

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

= \(\dfrac{-3}{3}+\dfrac{0}{5}\) = \(-1+0\) = \(-1\)

Giúp mik câu b nx lak đk

\(a.A=\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}=\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)=1+1=2\)

Vậy A=2

\(b.B=\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}=\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)

\(=\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)=-1+0=-1\)

Vậy B=-1

\(\dfrac{7}{21}+\dfrac{-9}{36}=\dfrac{1}{3}+\dfrac{-1}{4}=\dfrac{4}{12}+\dfrac{-3}{12}=\dfrac{1}{12}\)

\(\dfrac{-12}{18}+\dfrac{-21}{35}=\dfrac{-2}{3}+\dfrac{-3}{5}=\dfrac{-10}{15}+\dfrac{-9}{15}=\dfrac{-19}{15}\)

\(\dfrac{-18}{14}+\dfrac{15}{-21}=\dfrac{-9}{7}+\dfrac{-5}{7}=\dfrac{-14}{7}=-2\)

\(\dfrac{3}{21}+\dfrac{-6}{42}=\dfrac{1}{7}+\dfrac{-1}{7}=0\)

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

d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)

\(\sqrt{3-2\sqrt{2}}\)

a) \(\dfrac{4}{7}< \dfrac{6}{7}\)

\(\dfrac{15}{23}>\dfrac{12}{23}\)

\(\dfrac{45}{60}=\dfrac{3}{4}\)

b) \(\dfrac{8}{9}>\dfrac{11}{18}\)

\(\dfrac{13}{18}>\dfrac{2}{3}\)

\(\dfrac{27}{25}>1\)

a: 4/9+3/7=28/63+27/63=55/63

3/4+7/24=18/24+7/24=25/24

1/3+2/9+4/27=9/27+6/27+4/27=19/27

b: 5/6-3/8=20/24-9/24=11/24

7/15-11/30=14/30-11/30=3/30=1/10

2/3+1/6-7/12

=8/12+2/12-7/12

=3/12=1/4

c: 18/25*15/6=15/25*18/6=3*3/5=9/5

30/49:6/7=30/49*7/6=210/294=5/7

1/2*3/4:6/5=3/8*5/6=15/48=5/16

d: 8*3/5:12/5=24/5*5/12=2

4:9/5:10/3=4*5/9*3/10=2/3

Câu 1:

\(-\dfrac{5}{7}\cdot\dfrac{7}{-45}=\dfrac{1}{9}\) ; \(-\dfrac{3}{10}\cdot-30=9\) ; \(-12\cdot-\dfrac{2}{36}=\dfrac{2}{3}\)

Caau2;3;4;5,... tự bấm máy tính là ra

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)

\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)

\(=\dfrac{3}{x-y}\)