\(a,\dfrac{2}{3}-\dfrac{5}{6}+\dfrac{3}{4}\)

\(=(\dfrac{2}{3}+\dfrac{3}{4})-\dfrac{5}{6}\)

\(=\dfrac{7}{12}\)

\(b,\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{-4}{13}.\dfrac{6}{7}\)

\(=\dfrac{6}{7}.(\dfrac{8}{13}-\dfrac{4}{13})+(\dfrac{6}{13}.\dfrac{9}{7})\)

\(=(\dfrac{6}{7}.\dfrac{4}{13})+\dfrac{54}{91}\)

\(=\dfrac{24}{91}+\dfrac{54}{91}\)

\(=\dfrac{78}{91}=\dfrac{6}{7}\)

\(c,19.(-26)\)

\(=-494\)

a) `2/3 - 5/6 +3/4`

`=4/6-5/6+3/4`

`=-1/6+3/4`

`=-4/24+18/24`

`=14/24=7/12`

b) `6/7 .8/13 +6/13 .9/7 -4/13 .6/7`

`=6/7 .(8/13-4/13) + 6/13 .9/7`

`=6/7 . 4/13 + 6/13 . 9/7`

`= 6/13 .4/7 +6/13 .9/7`

`=6/13 . (4/7+9/7)`

`=6/13 . 13/7`

`=6/7`

c) `19.(-26)`

`=(1-20).(20+6)`

`=20+6-400-120`

`=-494`

a: =7+5/11-2-3/7-3-5/11

=2-3/7=11/7

b: =-3/5(5/7+3/7+6/7)

=-3/5*2=-6/5

c: =1/3(4/5+6/5)-4/3

=2/3-4/3=-2/3

d: =5/9(7/13+13-3/13)

=5/9*165/13=275/39

a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2

b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2

c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)

d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45

e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)

f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) .  ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)

g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)

h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)

i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\) 

                                     = \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\)  = \(\dfrac{-193}{200}\)

j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )

= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\) 

= -1+\(\dfrac{-13}{21}\)

= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)

_{Khôi nguyễn}

sao tính nổi:))

a \(\dfrac{53}{84}\)

b \(\dfrac{20}{63}\)

c \(\dfrac{1}{60}\)

a: \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{4}=\dfrac{56}{84}+\dfrac{60}{84}-\dfrac{63}{84}=\dfrac{53}{84}\)

b: \(=\dfrac{3}{7}-\dfrac{1}{9}=\dfrac{27}{63}-\dfrac{7}{63}=\dfrac{20}{63}\)

c: \(=\dfrac{2}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{3}=\dfrac{2}{8}\cdot\dfrac{1}{15}=\dfrac{1}{60}\)

Bài 1:

a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)

\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)

\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)

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

b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)

\(=3-\sqrt{6}+\sqrt{6}-1\)

=3-1=2

c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)

\(=\sqrt{15}+4-\sqrt{15}=4\)

d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

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

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

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

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

Bài 2:

Vẽ đồ thị:

Phương trình hoành độ giao điểm là:

\(\dfrac{1}{2}x-4=-3x+3\)

=>\(\dfrac{1}{2}x+3x=3+4\)

=>\(\dfrac{7}{2}x=7\)

=>x=2

Thay x=2 vào y=-3x+3, ta được:

\(y=-3\cdot2+3=-3\)

Vậy: (d1) cắt (d2) tại A(2;-3)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

a) 3/4+1/12 =5/6

b)11/4-6/7=53/28

c)2/9x8=16/9

d)8/21:4/7=2/3

a) 3/4+1/12 = 9/12 + 1/12= 10/12= 5/6

b)11/4-6/7 = 77/28 - 24/28 = 53/28

c)2/9x8= 16/9

d)8/21:4/7 = 8/21 x 7/4 = 2/3

a)5/6

b)53/28

c)16/9

d)2/3

a:=5/4-3/4=2/4=1/2

b: =2/3-2/5=10/15-6/15=4/15

c: =10/12-9/12=1/12

d: =4/3-1/4=16/12-3/12=13/12

\(a,=\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}\\ b,=\dfrac{30}{45}-\dfrac{18}{45}=\dfrac{12}{45}\\ c,=\dfrac{10}{12}-\dfrac{9}{12}=\dfrac{1}{12}\\ d,=1\dfrac{1}{12}\)

\(a,\dfrac{20}{16}-\dfrac{3}{4}=\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\ b,\dfrac{30}{45}-\dfrac{2}{5}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ c,\dfrac{10}{12}-\dfrac{3}{4}=\dfrac{5}{6}-\dfrac{3}{4}=\dfrac{10}{12}-\dfrac{9}{12}=\dfrac{1}{12}\\ d,\dfrac{12}{9}-\dfrac{1}{4}=\dfrac{4}{3}-\dfrac{1}{4}=\dfrac{16}{12}-\dfrac{3}{12}=\dfrac{13}{12}\)