a) 27 . 75 + 25 .27 - 150 = 27 . (75 + 25) - 150

                                       = 270 . 100 - 150

                                       = 27 000 - 150

                                       = 26 850

b) 3.5² - 16 : 2² = 12,25 - 16 : 4

                         = 12,25 - 4

                        = 8,25

c) 20 - [30 - (5 - 1)² ] = 20 - [30 - 4² ]

                                 = 20 - 30 - 16

                                 = (-10) - 16

                                = -26

d) 60 : {[(12 - 3) . 2] + 2} = 60 : {[9 . 2] + 2}

                                       = 60 : {18 + 2}

                                      = 60 : 20

                                      = 3

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

Bài 1: Thực hiện các phép tính sau:

\(a)\)Chưa rỏ đề

\(b)\)\(5025\div5-25\div5\)

\(=\)\(1005-5\)

\(=\)\(1000\)

\(c)\)\(218-180\div2\div9\)

\(=\)\(218-10\)

\(=\)\(208\)

\(d)\)\(\left(328-8\right)\div32\)

\(=\)\(320\div32\)

\(=\)\(10\)

Bài 1:

a) ( Tôi không nhìn rõ đầu bài )

b) 5025  :  5  -  25  :  5

=  ( 5025  -  25 )  :  5

=        5000   :   5

=       1000

c)   218  -  180  :  2  :  9

=     218   -    180  :  (  2  .  9 )

=      218    -    180   :  18

=       218    -    10

=       208

d)   (  328  -  8  )  :  32

=      320   :   32

=     10 

Bài 2: Thực hiện phép tính:

\(a)\)\(27.75+25.27-150\)

\(=\)\(27.\left(75+25\right)-150\)

\(=\)\(27.100-150\)

\(=\)\(2700-150\)

\(=\)\(2550\)

\(b)\)\(12\div\left\{400\div\left[500-\left(125+25.7\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-\left(125+175\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-300\right]\right\}\)

\(=\)\(12\div\left\{400\div200\right\}\)

\(=\)\(12\div2\)

\(=\)\(6\)

\(c)\)\(13.17-256\div16+14\div7-1\)

\(=\)\(221-16+2-1\)

\(=\)\(206\)

\(d)\)\(18\div3+182+3.\left(51\div17\right)\)

\(=\)\(6+182+3.3\)

\(=\)\(6+182+9\)

\(=\)\(197\)

Bài 1: 

a: =25+75=100

b: =60-17-43+12=12

c: =-2-18=-20

d: =-3+36-17=36-20=16

Bài 2: 

a: =-102

b: =-1000

c: =12x15=180

d: =21x(-10)=-210

B1

a 100

b 12

c -20

d 16

B2

a -102

b -1000

c 180

d -210

Bài 1:Tính

 a)(-12)+25+75+12 = 100

b)60+12+(-17)+(-43)=12
c)(-2)+(-87)+(-18)+87=-20
d)(-1)+(-2)+36+(-17)=16
Bài 2:Tính
b)8 . (-125)=-100
c)(-12) . (-15)=180
d)21 . (- 3)+(- 21 ) . 7=-210

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

a: =8/5-2/3=24/15-10/15=14/15

12/20-6/18=3/5-1/3=9/15-5/15=4/15

b: 18/21-12/27=6/7-7/9=54/63-49/63=5/63

8/12-20/50=2/3-2/5=10/15-6/15=4/15

\(a,\dfrac{8}{5}-\dfrac{18}{27}=\dfrac{8}{5}-\dfrac{2}{3}=\dfrac{14}{15}\\ \dfrac{12}{20}-\dfrac{6}{18}=\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{4}{15}\\ b,\dfrac{18}{21}-\dfrac{12}{27}=\dfrac{6}{7}-\dfrac{4}{9}=\dfrac{26}{63}\\ \dfrac{8}{12}-\dfrac{20}{50}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

27 . 75 + 25 . 27 - 150

= 27.(75+25)-150

= 27.100-150

= 2700 - 150

= 2550

20 - [ 30 - ( 5 - 1 )² ]

= 20 -[30-4² ]

= 20 - ( 14 )

= 6

27 . 75 + 25 . 27 - 150

= 27.(75+25)-150

= 27.100-150

= 2700 - 150

= 2550

20 - [ 30 - ( 5 - 1 )² ]

= 20 -[30-4² ]

= 20 - ( 14 )

= 6