\(\dfrac{15}{35}+\dfrac{7}{10}=\dfrac{3}{7}+\dfrac{7}{10}=\dfrac{30}{70}+\dfrac{49}{70}=\dfrac{79}{70}\)

\(\dfrac{27}{72}+\dfrac{11}{24}=\dfrac{3}{8}+\dfrac{11}{24}=\dfrac{9}{24}+\dfrac{11}{24}=\dfrac{20}{24}=\dfrac{5}{6}\)

\(\dfrac{6}{12}+\dfrac{7}{6}=\dfrac{1}{2}+\dfrac{7}{6}=\dfrac{3}{6}+\dfrac{7}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)

d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)

a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)

\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)

\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

cho mik xin bài giải với ạ

1: =72/90+65/90=137/90

2: =24/56-77/56=-53/56

3: =-7/10+4/5=1/10

4: =15/100-4/100=11/100

5: =4/6-5/6=-1/6

6: =10/40-15/40-76/40=-81/40

7: =-9/10+7/18

=-81/90+35/90=-46/90=-23/45

8: =27/90-55/90=-28/90=-14/45

9: =36/60-50/60-35/60=-49/60

10: =-4/9+5/6-3/8

=-32/72+60/72-27/72

=1/72

a: Ta có: \(\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\left(3-\frac34\right)\)

\(=\left(-\frac{6435}{9009}-\frac{7007}{9009}+\frac{7371}{9009}-\frac{7623}{9009}\right)\cdot\frac94\)

\(=-\frac{13694}{9009}\cdot\frac94=\frac{-6847}{2}\cdot\frac{1}{1001}=\frac{-6847}{2002}\)

b: \(\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac23\right)\)

\(=\left(\frac{12870}{27027}+\frac{14014}{27027}-\frac{14742}{27027}+\frac{15246}{27027}\right):\frac43\)

\(=\frac{27388}{27027}\cdot\frac34=\frac{6847}{9009}\)

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)