tính:
{[(32 + 1 ) . 10 - ( 8:2 + 6 )]: 2} + 55 - (10 : 5)3
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Tính các tổng sau:
1, S=1-2+3_4+..+25-26
S =-1+3-5+7-...-53+55 ( có 28 số hạng )
= (-1+3)+(-5+7)+...+(-53+55) ( có 28:2=14 nhóm )
= 2+2+...+2
= 2 . 14
= 28
{[( \(3^2\)+ 1 ) .10-(8:2 + 6 ) ] : 2 } + 55 - ( 10 : 5 )\(^3\)
= {[( 9 + 1 ) . 10 - ( 4 + 6 )] : 2 } + 55 - 2\(^3\)
= {[ 10 . 10 - 10 ] : 2 } + 55 - 8
= {[ 100 - 10 ] : 2 } + 47
= { 90 : 2 } + 47
= 45 + 47
= 92
Bài 1:
1: =15+37+52-37-17=52-2=50
2: =38-42+14-25+27+15=62-42+29=20+29=49
Bài 1: Bỏ ngoặc rồi tính
3) (21-32) - (-12+32)=21-32-(-12)-32=21-32+12-32=-31
4) (12+21) - (23-21+10)=12+21-23+21-10=21
5) (57-725) - (605-53)=57-725-605+53=-1220
6) (55+45+15) - (15-55+45)=55+45+15-15+55-45=55+55=110
Bài 2: Tính các tổng sau một cách hợp lí
1) (-37) + 14 + 26 + 37=(-37+37)+(14+26)=0+40=40
2) (-24) +6 + 10 + 24=(-24+24)+(6+10)=0+16=16
3) 15 + 23 + (-25) + (-23)=(15-25)+(23-23)=-10+0=-10
4) 60 + 33 + (-50) + (-33)=(60-50)+(33-33)=10+0=10
5) (-16) + (-209) + (-14) + 209=(-16-14)+(-209+209)=-30+0=-30
6) (-12) + (-13) + 36 + (-11)=(-11-12-13)+36=-36+36=0
B=2006 * 2008 -3 / 2005 + 2005 * 2008
B=(2005+1)* 2008 -3 / 2005 +2005 *2008
B=2005 * 2008 + 2008 -3 / 2005 +2005*2008
B=2005 * 2008 + 2005 / 2005 +2005 * 2008
B= 1
\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)
\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)
\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)
\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)
1: \(=\dfrac{7}{36}+\dfrac{32}{36}+\dfrac{-24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
2: \(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{3}{7}+\dfrac{4}{7}\)
\(=\dfrac{-9-2-7}{18}+1=-1+1=0\)
3: \(=\left(-\dfrac{10}{3}-\dfrac{1}{6}\right)+\dfrac{14}{10}\)
\(=-\dfrac{21}{6}+\dfrac{14}{10}=\dfrac{-7}{2}+\dfrac{14}{10}=\dfrac{-35+14}{10}=-\dfrac{21}{10}\)
4: \(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}=\dfrac{11}{25}\)
{ [ ( 32 + 1 ) . 10 - ( 8 : 2 + 6 ) ] : 2 } + 55 - ( 10 : 5 )
= { [ ( 9 + 1 ) . 10 - ( 8 : 2 + 6) ] : 2 } + 55 - ( 10 : 5 )
= { [ 10 . 10 - ( 8 : 2 + 6 ) ] : 2 } + 55 - ( 10: 5)
= { [ 10 . 10 - ( 4 + 6) ] : 2 } + 55 - (10 : 5)
= { [ 10 . 10 - 10 ] : 2 } + 55 - (10 : 5)
= { [ 10 .10 - 10 ] : 2 } + 55 - 2
= { [ 100 - 10 ] : 2 } + 55 - 2
= { 90 : 2 } + 55 - 2
= 45 + 55 - 2
= 100 - 2
= 98
\(\left\{\left[\left(3^2+1\right).10-\left(8:2+6\right)\right]:2\right\}+55-\left(10:5\right)^3\)
\(=\left\{\left[\left(9+1\right).10-\left(4+6\right)\right]:2\right\}+55-2^3\)
\(=\left\{\left[10.10-10\right]:2\right\}+55-8\)
\(=\left\{\left[100-10\right]:2\right\}+47\)
\(=\left\{90:2\right\}+47\)
\(=45+47\)
\(=92\)
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