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311<411; 411=(22)11=222
1714>1614; 1614=(24)14=256
Vì 222<256=> 411<1614
=>311<1714
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1 , \(\sqrt{7}\)-\(\sqrt{2}\)> 1
2 , \(^{\left(\sqrt{11}-x\right)^2}\)
Ta có : \(10.A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
\(10.B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1\)và \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\)nên \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)hay \(A>B\)
Vậy \(A>B\)
Có: \(a^3+b^3=c^3\Leftrightarrow\left(\frac{a}{c}\right)^3+\left(\frac{b}{c}\right)^3=1.\)
Đặt : \(\frac{a}{c}=x;\frac{b}{c}=y\). Suy ra \(0< x< 1;0< y< 1\).
Vì vậy: \(x^{2010}< x^3;y^{2010}< y^3.\)
Từ đó: \(x^{2010}+y^{2010}< x^3+y^3< 1\).
Suy ra: \(\left(\frac{a}{c}\right)^{2010}+\left(\frac{b}{c}\right)^{2010}< 1\)hay: \(a^{2010}+b^{2010}< c^{2010}.\)
Ta có: \(72^{45}-72^{44}=72^{44}\left(72-1\right)\)
\(72^{44}-72^{43}=72^{43}\left(72-1\right)\)
=> \(72^{44}\left(72-1\right)>72^{43}\left(72-1\right)\) hay \(72^{45}-72^{44}>72^{44}-72^{43}\)
T_i_c_k nha
Ta có:\(31^{13}< 32^{13}=\left(2^5\right)^{13}=2^{65}\)
\(65^{11}>64^{11}=\left(2^6\right)^{11}=2^{66}\)
Mà: \(2^{66}>2^{65}\Rightarrow65^{11}>2^{66}>2^{65}>31^{13}\)
Vậy: \(31^{13}< 65^{11}\)