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\(\dfrac{1}{2}\cdot\dfrac{5}{6}+\dfrac{2}{3}\cdot75\%\)

\(=\dfrac{1}{2}\cdot\dfrac{5}{6}+\dfrac{2}{3}\cdot\dfrac{3}{4}\)

\(=\dfrac{1\cdot5+2\cdot3}{12}=\dfrac{5+6}{12}=\dfrac{11}{12}\)

a)\(\left(-3\right).8.\left(-2\right).5=\left(-3\right).\left(-2\right).8.5\)

                                          \(=6.40=240\)

b)\(127-18.\left(5+6\right)=127-18.11\)

                                             \(=127-198=-71\)

c)\(125-\left(-75\right)+32-\left(48+32\right)=125+75+32-48-32\)

                                                                              \(\left(125+75\right)+\left(32-32\right)-48=200-48=152\)

d)\(3.\left(-4\right).2+2.\left(-5\right)-20=-12.2+2.\left(-5\right)-20\)

                                                             \(=\left(-12-5\right).2-20=-17.2-20=-140-20=-160\)

Chúc bạn học tốt

3 tháng 12 2023

a) (38 - 60) + (20 - 38)

= 38 - 60 + 20 - 38

= (38 - 38) + (-60 + 20)

= 0 - 40

= -40

b) 75 - (20 + 75)

= 75 - 20 - 75

= (75 - 75) - 20

= 0 - 20

= -20

c) 32 + (60 - 32)

= 32 + 60 - 32

= (32 - 32) + 60

= 0 + 60

= 60

d) (81 - 36) - (81 - 36)

= 81 - 36 - 81 + 36

= (81 - 81) + (-36 + 36)

= 0 + 0

= 0

e) (2 + 4 + 6 + 8) - (1 + 3 + 5 + 7)

= 2 + 4 + 6 + 8 - 1 - 3 - 5 - 7

= (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7)

= 1 + 1 + 1 + 1

= 4

f) (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + 8 + ... + 100)

= 1 + 3 + 5 + 7 + ... + 99 - 2 - 4 - 6 - 8 - ... - 100

= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + ... + (99 - 100)

= -1 - 1 - 1 - 1 - ... - 1 (50 chữ số 1)

= -50

3 tháng 12 2023

giúp mình với huhuhuhuhuhu

28 tháng 12 2023

Mai mình phải nộp bài nên  nhanh giúp mình ạ😅

27 tháng 6 2021

\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)

\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)

a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)

\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)

\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)

\(=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\)

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

21 tháng 12 2023

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:

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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)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)

`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)

`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)

`b)`

\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)

`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)

`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)

`c)`

\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)

`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)

23 tháng 7 2023

a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{6}-\dfrac{1}{3}\)

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

b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)

\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)

\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)

\(=\dfrac{85}{36}+\dfrac{-7}{12}\)

\(=\dfrac{16}{9}\)

c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)

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

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

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

a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)

\(=60-60\sqrt{2}+45\sqrt{3}\)

b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)

\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)

 

a: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-8+15}{18}:17\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{7}{18}\cdot\dfrac{1}{17}\)

\(=\dfrac{619}{918}\)

b: \(=\left(3-7+\dfrac{1}{4}\right)\cdot\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{-15}{4}\cdot\dfrac{13}{12}=\dfrac{-195}{48}=\dfrac{-65}{16}\)

c: \(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}-\dfrac{1}{4}\)

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

d: \(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)\)

\(=\dfrac{5}{16}:\dfrac{-3}{16}=\dfrac{-5}{3}\)