a) 143 - 12 . 5 = 143 - 60 = 83

b) 27 . 8 - 6 : 3  = 216 - 2 = 214

c) 36 - 12 : 4 . 3 + 17 = 36 - 3 . 3 + 17 = 36 - 9 + 17 = 27 + 17 = 44

a) 143 - 12 . 5 = 143 - 60 = 83

b) 27 . 8 - 6 : 3  = 216 - 2 = 214

c) 36 - 12 : 4 . 3 + 17 = 36 - 3 . 3 + 17 = 36 - 9 + 17 = 27 + 17 = 44

\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)

a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)

b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)

c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)

a) (-12). (7 - 72) - 25. (55 - 43) 

= (-12). (- 65) - 25. 12 

= 12. 65 – 12. 25

= 12. (65 - 25) 

= 12. 40 

= 480

b) (39 - 19) : (- 2) + (34 - 22). 5 

= 20 : (- 2) + 12. 5 

= - 10 + 60

= 60 - 10 

= 50.

ai lm tr mình cho tích

a) Với a = 3,05 thì ta có:

\(a\times2,46\) \(=3,05\times2,46=7,503\)

b) Với \(a=\dfrac{15}{8}\) thì ta có:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\) \(=\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\) \(=\left(\dfrac{10}{12}+\dfrac{7}{12}\right)\times\dfrac{8}{15}\) \(=\dfrac{17}{12}\times\dfrac{8}{15}\) \(=\dfrac{34}{45}\)

a) Thay a=3,05 vào 2,46a, ta được:

\(2.46\cdot3.05=7.503\)

b) Thay \(a=\dfrac{15}{8}\) vào biểu thức \(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\), ta được:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\)

\(=\dfrac{17}{12}\cdot\dfrac{8}{15}=\dfrac{34}{45}\)

Sửa đề: A=7/12-9/20+11/30-13/42+15/56

A=1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8

=1/3+1/8=11/24

a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

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

                                              = \(2\left(\sqrt{2}-1\right)\)

b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)

Ta có:

B² = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)

     = \(12-2\sqrt{36-20}\)

     = \(12-8\)

     = \(4\)

\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2

Vậy B = 2

c) C = \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\) (C > 0)

Ta có: 

C² = \(17+12\sqrt{2}+2\sqrt{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}+\left(17-12\sqrt{2}\right)\)

     = \(34+2\sqrt{289-288}\)

     = \(34+2\)

     = \(36\)

\(\Rightarrow C=\pm6\) nhưng vì C > 0 nên C = 6

`#lv`

`A=(-1)+(-5)+(-9)+...+(-101)`

`=-(1+5+9+...+101)`

Số số hạng là : 

`[101-(-1)]:4+1=26(` số hạng `)`

Tổng là : 

`[(-101)+(-1)]xx26:2=-1326`

Vậy `A=-1326`

__

`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`

`=((-5)/17+(-12)/17).8/19-11/19`

`=-1.8/19-11/19`

`=-8/19-11/19`

`=-8/19+(-11)/19`

`=-19/19`

`=-1`

__

`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`

`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`

`=2(1-1/2021)`

`=2. (2021/2021-1/2021)`

`=2. 2020/2021`

`=4040/2021`