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\(Ư\left(144\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\right\}\)
\(Ư\left(324\right)=\left\{1,2,3,4,6,9,12,18,27,36,54,81,108,162,324\right\}\)
\(Ư\left(576\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,32,36,48,64,72,96,144,192,288,576\right\}\)
\(Ư\left(1024\right)=\left\{1,2,4,8,16,32,64,128,256,512,1024\right\}\)
\(Ư\left(1296\right)=\left\{1,2,3,4,6,8,9,12,16,24,27,36,48,54,72,81,108,144,162,216,324,432,648,1296\right\}\)
Cách 1:
a) \(12:6 = 2\)
b) \(24:\left( { - 8} \right)=-(24:8)=-3\)
c) \(\left( { - 36} \right):9=-(36:9)=-4\)
d) \(\left( { - 14} \right):\left( { - 7} \right)=14:7=2\)
Cách 2:
a) Ta có \(12 = 6.2\) nên \(12:6 = 2\).
b) Ta có \(24 = \left( { - 8} \right).\left( { - 3} \right)\)\( \Rightarrow 24:\left( { - 8} \right) = \left( { - 3} \right)\).
c) Ta có \(\left( { - 36} \right) = 9.\left( { - 4} \right)\) nên \(\left( { - 36} \right):9 = \left( { - 4} \right)\).
d) Ta có \(\left( { - 14} \right) = \left( { - 7} \right).2\) nên \(\left( { - 14} \right):\left( { - 7} \right) = 2\)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
\(a.\)
\(\dfrac{3}{10}:\left(-\dfrac{2}{3}\right)=\dfrac{3}{10}\cdot\dfrac{-3}{2}=-\dfrac{9}{20}\)
\(b.\)
\(\left(-\dfrac{7}{12}\right):\left(-\dfrac{5}{6}\right)=\left(-\dfrac{7}{12}\right)\cdot\left(-\dfrac{6}{5}\right)=\dfrac{\left(-7\right)\cdot\left(-6\right)}{12\cdot5}=\dfrac{7}{10}\)
\(c.\)
\(\left(-15\right):-\dfrac{9}{10}=\left(-15\right)\cdot-\dfrac{10}{9}=\dfrac{150}{9}=\dfrac{50}{3}\)
a) \(\dfrac{3}{10}:\dfrac{-2}{3}=\dfrac{3}{10}.\dfrac{-3}{2}=\dfrac{3.-3}{10.2}=\dfrac{-9}{20}\)
b) \(\dfrac{-7}{12}:\dfrac{-5}{6}=\dfrac{-7}{12}.\dfrac{-6}{5}=\dfrac{-7.-6}{12.5}=\dfrac{7}{10}\)
c)\(-15:\dfrac{-9}{10}=-15.\dfrac{-10}{9}=\dfrac{-15.-10}{9}=\dfrac{50}{3}\)
`#3107.101107`
`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`
`= -9 + [-54 \div (-256 + 7) * 4]`
`= -9 + [-54 \div (-249) * 4]`
`= -9 + (18/83 * 4)`
`= -9 + 72/83`
`= -675/83`
______
`31 * (-18) + 31 * (-81) - 31`
`= 31 * (-18 - 81 - 1)`
`= 31 * (-100)`
`= -3100`
___
`(-12) * 47 + (-12) * 52 + (-12)`
`= (-12) * (47 + 52 + 1)`
`= (-12) * 100`
`= -1200`
___
`13 * (23 + 22) - 3 * (17 + 28)`
`= 13 * 45 - 3 * 45`
`= 45 * (13 - 3)`
`= 45 * 10`
`= 450`
____
`-48 + 48 * (-78) + 48 * (-21)`
`= 48 * (-1 - 78 - 21)`
`= 48 * (-100)`
`= -4800`
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
a)(2;3;4;5;.....;3207
b)tap hop B la tập hợp con của tập hợp A
Bài 1:
a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)
\(=12+\left\{45-\left[36:3^2\right]\right\}\)
\(=12+\left\{45-\left[36:9\right]\right\}\)
\(=12+\left\{45-4\right\}\)
\(=12+41\)
\(=53\)
b.\(24:\left[48-\left(42:7\right)^2\right]\)
\(=24:\left[48-6^2\right]\)
\(=24:\left[48-36\right]\)
\(=24:12=2\)
Bài 2 :
C1 : { 0;1;2;3;4;5;6;7;8;9;10}
C2 : \(\left\{x\in N|x\le10\right\}\)
pha ngoac ra la dc
con bai 2 C1 {1:2:3:4:5:6:7:8:9:0}
C2 {nt huoc N /n<10}