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\(A=\dfrac{-10}{3}+\dfrac{19}{6}\cdot\dfrac{7}{5}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{10}\cdot\dfrac{4}{3}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{7}{10}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{3}\cdot\dfrac{4}{10}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\left(\dfrac{7}{10}-\dfrac{1}{10}+\dfrac{4}{10}\right)\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{10}{10}=\dfrac{-10}{3}+\dfrac{19}{3}\)
\(=\dfrac{9}{3}=3\)
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a, \(10^m-1⋮19,19⋮19\)
\(\Rightarrow\left(10^m-1\right)\left(10^m+1\right)+19⋮19\)
\(\Rightarrow10^{2m}-1+19⋮19\Rightarrow10^{2m}+18⋮19\)
\(b,\)Ta có : \(3+3^2+3^3+3^4+...+3^{23}+3^{24}+3^{25}\)
\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{23}+3^{24}+3^{25}\right)\)
\(=3+3\left(3+3^2+3^3\right)+...+3^{22}\left(3+3^2+3^3\right)\)
\(=3+3.39+...+3^{22}.39\)
\(=3+39\left(3+...+3^{22}\right)\)
Suy ra : B chia 39 dư 3
Vậy : B không chia hết cho 39
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A = -10/3 + 19/6.7/5 - 19/3.1/10 + 19/10.4/3
A = 1/30.(-10.10 + 19.7 - 19.1 + 19.4)
A = 1/30.(-100 + 19.11 - 19)
A = 1/30.(-100 + 19.10)
A = 1/30.(-100 + 190)
A = 1/30.90
A = 3
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\(\frac{2^{19}\times27^3+15\times4^9\times9^4}{6^9\times2^{10}+12^{10}}\)
\(=\frac{2^{19}\times\left(3^3\right)^3+5\times3\times\left(2^2\right)^9\times\left(3^2\right)^4}{\left(2\times3\right)^9\times2^{10}+\left(3\times4\right)^{10}}\)
\(=\frac{2^{19}\times3^9+3\times5\times2^{18}\times3^8}{3^9\times2^9\times2^{10}+3^{10}\times4^{10}}\)
\(=\frac{2^{19}\times3^9+5\times2^{18}\times3^9}{3^9\times2^{19}+3^{10}\times\left(2^2\right)^{10}}\)
\(=\frac{2^{18}\times3^9\times\left(2+5\right)}{3^9\times2^{19}+3^{10}\times2^{20}}\)
\(=\frac{2^{18}\times3^9\times7}{2^{19}\times3^9\times\left(1+3\times2\right)}\)
\(=\frac{7}{2\times\left(1+6\right)}\)
\(=\frac{7}{2\times7}\)
\(=\frac{1}{2}\)
A = 2^19.27^3+15.4^9.9^4 / 6^9.2^10+12^10
= 2^19.3^9 + 5.2^18.3^9 / 3^9.2^19 + 2^20.3^10
= 2^18.3^9 ( 2 + 5 ) / 2^19.3^9.(1 + 2.3)
= (2 + 5) / 2(1 + 6)
= 7 / 2.7
= 1/2
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\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
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\(C=3^{n+2}-2^{n+2}+3^n-2^n\)
\(C=\left(3^{n+2}-2^{n+2}\right)+\left(3^n-2^n\right)\)
\(\Rightarrow C=1^{n+2}+1^n\) (với n \(\in\)N*)
Ta có công thức Cơ số có tận cùng bằng 1 thì mũ lên bao nhiêu cũng bằng 1.(với n \(\in\)N*)
Vì n \(\in\)N* \(\Rightarrow C=1^{n+2}+1^n=\left(...1\right)+\left(...1\right)=\left(...2\right)\)
11×4+24×10+310×19+419×3111×4+24×10+310×19+419×31
=13×(31×4+64×10+910×19+1219×31)=13×(31×4+64×10+910×19+1219×31)
=13×(1−14+16−110+110−119+119−131)=13×(1-14+16-110+110-119+119-131)
=13×(1−131)=13×(1-131)
=13×3031=13×3031
=1031