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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1 2024
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 2024
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 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
17 tháng 8 2016
a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)
\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)
\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)
b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)
\(=-\frac{5}{72}:\frac{3.7}{8.9}\)
\(=-\frac{5}{72}:\frac{7}{24}\)
\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)
17 tháng 8 2016
a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)
b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)
c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)
\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)
\(=\frac{-23}{25}+\frac{-27}{25}\)
\(=\frac{-50}{25}=-2\)
1 tháng 12 2021
Câu 1 : Chọn ý ko đúng trong các câu sau
A. (-15) + 23 < 12
B.(-21) + (-13) = -33
C.16+(-25)>3
D.(-13)+(-17)<-10
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}
A.\(-|-13|+\left(-23\right)\)
= -13+ (-23)
=-36
B.\(-|13|+|-25|+|12|\)
=-13+25+12
=24