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TG
6 tháng 8 2020
\(A=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(=98.42-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(=98.42-\left[50\left(10:2+9\right)\right]\)
\(=98.42-\left(50.14\right)\)
\(=4116-700=3416\)
\(B=-80-\left[-130-\left(12-4\right)^2\right]+2008^0\)
\(=-80-\left(-130-8^2\right)+1\)
\(=-80-\left(-130-64\right)+1\)
\(=-80+130+64+1\)
\(=115\)
\(C=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(=1024:16+140:\left(38+32\right)-7^2\)
\(=64+140:70-49\)
\(=64+2-49=17\)
\(D=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0\)
\(=0\)
\(E=100+98+96+....+4+2-97-95-....-3-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(2-1\right)+\left(1-0\right)\)
\(=100+1+1+...+1+1\)
Vì lập được 49 cặp nên sẽ có 49 số 1
\(\Rightarrow E=100+1.49=100+49=149\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
8 tháng 11 2021
\(a,=2^3\left(17-14\right)=2^3\cdot3=24\\ b,=80-\left(130-8^2\right)=80-\left(130-64\right)=80-66=14\)
mời bạn đừng thi thì tốt hơn 
29 tháng 3 2020
a. (+18) + (+2)= 20
b. (-3)+13= 11
c. (-12)+(-21)=-33
d. (-30)+(-23)=-53
e. -52+102=50
f. 88+(-23)=65
g. 13+|-13|=26
h. -43-26=-69
k. (-89)-9=-98
m. -56+|-32|=-24
n. 40 -|-14|=26
o. |-4|+|+15|=19
p. |30|-|-17|=13
E
25 tháng 2 2020
a) 100 + (+430) + 2145 + (-530)
= 100 + [ (+430) + (-530) ] + 2145
= 100 + (-100) +2145
= 0 + 2145 = 2145
b) (-12) . 15 . 8 . (-4)
= [ (-12) . 8 ] . [ 15. (-4)]
= (-96) . (-60) = 5760
c) (+12) . 13 + 13 . (-22)
= 13. [ (+12) + (-22) ]
= 13. (-10) = -130
d) { [14 : (-2)] + 7 } : 2012
= { [ (-7) + 7 } : 2012
= 0 : 2012 = 0
e) -23 . 63 + 23 . 21 - 58 . 23
= 23 . -63 + 23 . 21 - 58.23
= 23. ( -63 + 21 - 58 )
= 23 . -100 = -2300
Hok tốt !
25 tháng 2 2020
a) 100 + (+430) + 2145 + (-530)
=100+[430+(-530)]+2145
=100+(-100)+2145
=0+2145
=2145
b) (-12) . 15 . 8 . (-4)
=(-12.8).[15.(-4)]
=-90.(-60)
=5760
c) (+12) . 13 + 13 . (-22)
=13.[12+(-22)]
=13.(-10)
=-130
d) { [14 : (-2)] + 7 } : 2012
={-7+7}:2012
=0:2012
=0
e) -23 . 63 + 23 . 21 - 58 . 23
=23.(-63)+23.21-58.23
=23.(-63+21-58)
=23.(-100)
=-2300
5 tháng 2 2020
a, 43 + ( 9 - 21 ) = 317 - ( x + 317 )
43 + ( -12 ) = 317 - x - 317
43 - 12 = 317 - 317 - x
-x = 31
x = -31
b, (15-x) + (x-12) = 7- (-5 + x)
15-x+x-12 = 7+5-x
15-12 = 12-x
3 = 12-x
x = 12-3
x = 9
c, x - { 57- [42+ (-23 - x)]} = 13- {47+ [25- (32-x)]}
x - [57- (42-23-x)] = 13- [47+ (25-32+x)]
x - [57- (19-x)] = 13- [47+ (x-7)]
x - (57-19+x) = 13- (47+x-7)
x - (38+x) = 13- (40+x)
x-38-x = 13-40-x
x = 13-40+38
x = 11
4 tháng 2 2020
a) \(43+\left(9-21\right)=317-\left(x+317\right)\\ 43+9-21=317-x-317\\ 52-21=\left(317-317\right)-x\\ 31=-x\\ x=-31\)Vậy x = -31
b) \(\left(15-x\right)+\left(x-12\right)=7-\left(-5+x\right)\\ 15-x+x-12=7+5-x\\ \left(x-x\right)+\left(15-12\right)=12-x\\ 3=12-x\\ x=9\)Vậy x = 9
c) \(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\\ x-\left\{57-\left[42+\left(-23\right)-x\right]\right\}=13-\left\{47+\left[25-32+x\right]\right\}\\ x-\left\{57-42+23+x\right\}=13-\left\{47+25-32+x\right\}\\ x-57+42-23-x=13-47-25+32-x\\ -57+42-23=-34-25+32-x\\ -15-23=-59+32-x\\ -38=-27-x\\ x=11\)Vậy x = 11
d) \(-7+\left|x-4\right|=-3\\ \left|x-4\right|=4\\ \Rightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\)Vậy \(x\in\left\{8;0\right\}\)
e) \(13-\left|x+5\right|=13\\ \left|x+5\right|=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)Vậy x = -5
g) \(\left|x-10\right|-\left(-12\right)=4\\ \left|x-10\right|=-8\\ \Rightarrow x\in\varnothing\left(\text{vì }\left|x-10\right|\ge0\text{với mọi }x\right)\)Vậy \(x\in\varnothing\)
h) \(\left|x+2\right|< 5\\ 0\le\left|x+2\right|< 5\\ \Rightarrow\left|x+2\right|\in\left\{1;2;3;4\right\}\\ \Rightarrow x+2\in\left\{1;-1;2;-2;3;-3;4;-4\right\}\\ \Rightarrow x\in\left\{-1;-3;0;-4;1;-5;2;-6\right\}\)Vậy \(x\in\left\{-1;-3;0;-4;1;-5;2;-6\right\}\)
5 tháng 2 2020
a) 43+(9−21)=317−(x+317)43+9−21=317−x−31752−21=(317−317)−x31=−xx=−3143+(9−21)=317−(x+317)43+9−21=317−x−31752−21=(317−317)−x31=−xx=−31Vậy x = -31
b) (15−x)+(x−12)=7−(−5+x)15−x+x−12=7+5−x(x−x)+(15−12)=12−x3=12−xx=9(15−x)+(x−12)=7−(−5+x)15−x+x−12=7+5−x(x−x)+(15−12)=12−x3=12−xx=9Vậy x = 9
c) x−{57−[42+(−23−x)]}=13−{47+[25−(32−x)]}x−{57−[42+(−23)−x]}=13−{47+[25−32+x]}x−{57−42+23+x}=13−{47+25−32+x}x−57+42−23−x=13−47−25+32−x−57+42−23=−34−25+32−x−15−23=−59+32−x−38=−27−xx=11x−{57−[42+(−23−x)]}=13−{47+[25−(32−x)]}x−{57−[42+(−23)−x]}=13−{47+[25−32+x]}x−{57−42+23+x}=13−{47+25−32+x}x−57+42−23−x=13−47−25+32−x−57+42−23=−34−25+32−x−15−23=−59+32−x−38=−27−xx=11Vậy x = 11
d) −7+|x−4|=−3|x−4|=4⇒[x−4=4x−4=−4⇒[x=8x=0−7+|x−4|=−3|x−4|=4⇒[x−4=4x−4=−4⇒[x=8x=0Vậy x∈{8;0}x∈{8;0}
e) 13−|x+5|=13|x+5|=0⇒x+5=0⇒x=−513−|x+5|=13|x+5|=0⇒x+5=0⇒x=−5Vậy x = -5
g) |x−10|−(−12)=4|x−10|=−8⇒x∈∅(vì |x−10|≥0với mọi x)|x−10|−(−12)=4|x−10|=−8⇒x∈∅(vì |x−10|≥0với mọi x)Vậy x∈∅x∈∅
h) |x+2|<50≤|x+2|<5⇒|x+2|∈{1;2;3;4}⇒x+2∈{1;−1;2;−2;3;−3;4;−4}⇒x∈{−1;−3;0;−4;1;−5;2;−6}|x+2|<50≤|x+2|<5⇒|x+2|∈{1;2;3;4}⇒x+2∈{1;−1;2;−2;3;−3;4;−4}⇒x∈{−1;−3;0;−4;1;−5;2;−6}Vậy x∈{−1;−3;0;−4;1;−5;2;−6}
6 tháng 12 2015
1.Tính 2.Tính
a)20 a)-73
b)-25 b)-30
c)-140 c)-18
3.So sánh
a) A>B
b) A=B
4. So sánh
a)A<B
b)A>B
c) A<B
Tick cho mình nha!!!
bạn ei, con c chép sai đề rồi
ukm , sai đề