\(9^{100}và3^{200}=3^{200}và3^{200}\\ \Rightarrow3^{200}=3^{200}\\ \Rightarrow9^{100}=3^{200}.\\ 5^{23}và125^3=5^{23}và5^9\\ \Rightarrow5^{23}>5^9\\ \Rightarrow5^{23}>5^3.\)

\(3.38.8+4.37.6+2.25.12\\ =24.38+24.37+24.25\\ =24.\left(38+37+25\right)\\ =24.100\\ =2400.\)

\(x\times2+x\times8=620\\ x\times\left(2+8\right)=620\\ x\times10=620\\ x=620:10\\ x=62.\)

Bổ sung:

Vì n là số tự nhiên

\(\Rightarrow n=0.\)

Vì \(12⋮2n+3\) nên

\(2n+3\inƯ\left(12\right)=\left\{1;2;3;4;6;12\right\}\)

Lập bảng:

Vậy \(n\in\left\{-1;-\dfrac{1}{2};0;\dfrac{1}{2};\dfrac{3}{2};\dfrac{9}{2}\right\}\)

\(39,52-23,48+20,48-26,52\\ =\left(39,52-26,52\right)+\left(-23,48+20,48\right)\\ =13-3\\ =10.\)

\(\overline{21\cdot}\)

Để \(\overline{21\cdot}\) là hợp số thì * \(\in\left\{0;2;3;4;5;6;8;9\right\}\)

\(a,38+41+117+159+62\\ =\left(38+62\right)+\left(41+159\right)+117\\ =100+200+117\\ =300+117\\ =417.\\ b,73+86+914+3032\\ =73+1000+3032\\ =1073+3032\\ =5150.\\ c,314.67+314.16+659.83\\ =314.\left(67+16\right)+659.83\\ =314.83+659.83\\ =83.\left(314+659\right)\\ =83.937\\ =?\)

Bn xem lại phần \(c,\) .

\(d,42.53+47.156-47.114\\ =42.53+47.\left(156-114\right)\\ =42.53+47.42\\ =42.\left(53+47\right)\\ =42.100\\ =4200.\)

Ta có: \(20=2^2.5\\ 2=2\)

\(\Rightarrow BCNN\left(20,2\right)=2^2.5=20.\\ \Rightarrow x=20.\)