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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
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
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
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
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
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
\(a.27.75+25.27-150\)
\(=27.\left(75+25\right)-150\)
\(=27.100-150\)
\(=2700-150\)
\(=2550\)
a, 27.75+25.27-150 b,3.52-16:23 d, 60 : {[( 12-3 ).2] +2}
= 27.(75+25)-150 = 3.25-16:4 = 60 : {[ 9.2] + 2
= 27. 100 - 150 = 75 - 4 = 60 : 18 + 2
= 2700 - 150 = 71 = 60 : 20
= 2550 = 3
Tk mik nha