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a) Ta có: \(14-\left(5-8\right)^3+\left(-3\right)\cdot5\)
\(=14+27-15=26\)
b) Ta có: \(-7\cdot15+7\cdot\left(-35\right)+\left(-1\right)^{2019}\)
\(=-7\left(15+35\right)-1=-350-1=-351\)
c) Ta có: \(-18\cdot32+\left(-18\right)\cdot45+77\cdot\left(-32\right)-77\cdot50\)
\(=-18\left(32+45\right)+77\left(-32-50\right)\)
\(=-18\cdot77+77\cdot\left(-82\right)=77\left(-18-82\right)=77\cdot\left(-100\right)=-7700\)
d) Ta có: \(104-4\cdot\left[-5\cdot8+\left(7-10\right)^3\right]\)
\(=104-4\left[-40-27\right]=104-4\cdot\left(-67\right)=104+268=372\)
Bài 1:
a) Ta có: \(1000\cdot25\cdot4\)
\(=10^3\cdot10^2\)
\(=10^5\)
b) Ta có: \(32\cdot24\cdot243\)
\(=2^5\cdot2^3\cdot3\cdot3^5\)
\(=2^8\cdot3^6\)
\(=6^6\cdot2^2\)
\(=\left(6^3\right)^2\cdot2^2\)
\(=\left(6^3\cdot2\right)^2=432^2\)
c) Ta có: \(8\cdot9\cdot10\cdot125\)
\(=2^3\cdot3^2\cdot2\cdot5\cdot5^3\)
\(=2^4\cdot3^2\cdot5^4\)
\(=\left(10^2\right)^2\cdot3^2\)
\(=\left(10^2\cdot3\right)^2\)
\(=300^2\)
d) Ta có: \(6^{15}\cdot6^{21}\cdot36\)
\(=6^{15}\cdot6^{21}\cdot6^2\)
\(=6^{38}\)
e) Ta có: \(\frac{5^{2019}}{5^{2000}}\cdot625\)
\(=5^{19}\cdot5^4=5^{23}\)
Bài 2:
a) Ta có: 57<75
nên \(29^{57}< 29^{75}\)
\(C=5^{2018}+\frac{1}{5^{2017}+1}=\left(5^{2017}+1\right)+\frac{1}{5^{2017}+1}\)
\(D=5^{2018}+\frac{1}{5^{2018}+1}=\left(5^{2017}+1\right)+\left(1+\frac{1}{5^{2017}+2}\right)\)
Do \(\frac{1}{5^{2017}+1}< 1+\frac{1}{5^{2017}+2}\)
Nên \(C< D\)
Ta có : C = \(\frac{5^{2018}+1}{5^{2017}+1}\)
=> \(\frac{C}{5}=\frac{5^{2018}+1}{5^{2018}+5}=1-\frac{4}{5^{2018}+5}\)
Lại có D = \(\frac{5^{2019}+1}{5^{2018}+1}\)
=> \(\frac{D}{5}=\frac{5^{2019}+1}{5^{2019}+5}=1-\frac{4}{5^{2019}+5}\)
Vì \(\frac{4}{5^{2018}+5}>\frac{4}{5^{2019}+5}\Rightarrow1-\frac{4}{5^{2018}+5}< 1-\frac{4}{5^{2019}+5}\Rightarrow\frac{C}{5}< \frac{D}{5}\Rightarrow C< D\)
Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Ta có: \(\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2019}-1+11}{10^{2020}-1+11}=\frac{10^{2019}+10}{10^{2020}+10}=\frac{10.\left(10^{2018}+1\right)}{10.\left(10^{2019}+1\right)}=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2018}+1}{10^{2019}+1}\)
Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)
\(B=\frac{10^{2018}+1}{10^{2019}+1}\)
Dễ thấy \(A< 1\)
Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với m>0
Vậy \(A=\frac{10^{2019}-1}{10^{2020}-1}< \frac{\left[10^{2019}-1\right]+11}{\left[10^{2020}-1\right]+11}=\frac{10^{2019}+10}{10^{2020}+10}\)
\(A< \frac{10\left[10^{2018}+1\right]}{10\left[10^{2019}+1\right]}=\frac{10^{2018}+1}{10^{2019}+1}=B\)
Do đó : A<B
\(C=\frac{29^{2019}+1}{29^{2019}-1};D=\frac{29^{2019}-1}{29^{2019}-3}\)
Đặt phân số trung gian là \(\frac{29^{2019}-1}{29^{2019}-1}\)
\(\Rightarrow C>\frac{29^{2019}-1}{29^{2019}-1}>D\)
Từ đó suy ra C > D.
Ta có : \(C=\frac{29^{2019}+1}{29^{2019}-1}>\frac{29^{2019}-1}{29^{2019}-1}\)( tử lớn hơn )
\(D=\frac{29^{2019}-1}{29^{2019}-3}< \frac{29^{2019}-1}{29^{2019}-1}\)( mẫu lớn hơn thì nhỏ hơn )
\(\Rightarrow C>\frac{29^{2019}-1}{29^{2019}-1}>D\)
\(\Leftrightarrow C>D\)