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a) \(4.8^6.2.8^3\)
\(=2^2.\left(2^3\right)^6.2.\left(2^3\right)^3\)
\(=2^2.2^{18}.2.2^9\)
\(=2^{2+18+1+9}\)
\(=2^{30}\)
______
b) \(12^2.2.12^3.6\)
\(=12^2.12^3.2.6\)
\(=12^2.12^3.12\)
\(=12^{2+3+1}\)
\(=12^6\)
c) \(6^3.2.6^4.3\)
\(=6^3.6^4.2.3\)
\(=6^3.6^4.6\)
\(=6^{3+4+1}\)
\(6^8\)
a) \(4\cdot8^6\cdot2\cdot8^3\)
\(=2^2\cdot\left(2^3\right)^6\cdot2\cdot\left(2^3\right)^3\)
\(=2^2\cdot2^{18}\cdot2\cdot2^9\)
\(=2^{30}\)
b) \(12^2\cdot2\cdot12^3\cdot6\)
\(=12^2\cdot12\cdot12^3\)
\(=12^6\)
c) \(6^3\cdot2\cdot6^4\cdot3\)
\(=6^3\cdot6\cdot6^4\)
\(=6^8\)
3n + 4 = 3n - 6 + 10
= 3(n - 2) + 10
Để (3n + 4) ⋮ (n - 2) thì 10 ⋮ (n - 2)
⇒ n - 2 ∈ Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}
⇒ n ∈ {-8; -3; 0; 1; 3; 4; 7; 12}
Mà n là số tự nhiên
⇒ n ∈ {0; 1; 3; 4; 7; 12}
\(A=2+2^2+2^3+...+2^{260}\)
\(A=2\left(1+2\right)+2^2\left(1+2\right)+2^3\left(1+2\right)+...+2^{259}\left(1+2\right)\)
\(A=2.3+2^2.3+2^3.3+...+2^{259}.3\)
\(A=3\left(2+2^2+2^3+...+2^{259}\right)⋮3\left(1\right)\)
\(A=\left(2+2^2+2^3\right)+...+\left(2^{258}+2^{259}+2^{260}\right)\)
\(A=2.\left(1+2+2^2\right)+...+2^{258}.\left(1+2+2^2\right)\)
\(A=2.7+...+2^{258}.7\Rightarrow A=7\left(2+...+2^{258}\right)⋮7\left(2\right)\)
\(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{257}+2^{258}+2^{259}+2^{260}\right)\)
\(A=2.\left(1+2+2^2+2^3\right)+...+2^{257}.\left(1+2+2^2+2^3\right)\)
\(A=2.15+...+2^{257}.15\Rightarrow A=15\left(2+...+2^{257}\right)⋮5\left(15⋮5\right)\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow dpcm\)
a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
Bài 3:
a) \(159-\left(25-x\right)=43\)
\(\Rightarrow25-x=159-43\)
\(\Rightarrow25-x=116\)
\(\Rightarrow x=25-116\)
\(\Rightarrow x=-91\)
b) \(\left(79-x\right)-43=-\left(17-52\right)\)
\(\Rightarrow\left(79-x\right)-43=-\left(-35\right)\)
\(\Rightarrow79-x=35+43\)
\(\Rightarrow79-x=78\)
\(\Rightarrow x=79-78\)
\(\Rightarrow x=1\)
c) \(-\left(-x+13-142\right)+18=55\)
\(\Rightarrow-\left(-x+13-142\right)=55-18\)
\(\Rightarrow x-13+142=37\)
\(\Rightarrow x+129=37\)
\(\Rightarrow x=37-129\)
\(\Rightarrow x=-92\)
\(5,2863< \overline{5,a8b1}< 5,3841\)
=>\(2863< \overline{a8b1}< 3841\)
=>\(\left(a,b\right)\in\left\{\left(2;7\right);\left(2;8\right);\left(2;9\right);\left(3;0\right);\left(3;1\right);\left(3;2\right);\left(3;3\right)\right\}\)
= 23,18 - 3,18 - 4,17 - 5,83 + 51,54 + 8,46
= 20 - 10 + 60
= 70
1) Ta có: \(1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=-1\cdot10=-10\)
2) Ta có: \(1-2+3-4+...+99-100\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=-1\cdot50=-50\)
mình cảm ơn ạ .