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\(a,101^2=101.\left(100+1\right)=10100+101=10201.\\ b,75^2-50.75+25^2\\ =75.\left(75-50\right)+25^2\\ =75.25+25^2\\ =25.\left(75+25\right)\\ =25.100\\ =2500.\)
\(c,103.97\\ =\left(100+3\right).97\\ =9700+291\\ =9991\)
a.
\(\left(0,25\right)^3\times32\)
\(=\left(0,25\right)^3\times2^5\)
\(=\left(0,25\right)^3\times2^3\times2^2\)
\(=\left(0,25\times2\right)^3\times4\)
\(=\left(0,5\right)^3\times4\)
\(=0,125\times4\)
\(=0,5\)
b.
\(\left(-0,125\right)^3\times80^4\)
\(=\left(-0,125\right)^3\times80^3\times80\)
\(=\left(-0,125\times80\right)^3\times80\)
\(=\left(-10\right)^3\times80\)
\(=-1000\times80\)
\(=-80000\)
c.
\(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}\times\left(3^2+3-1\right)\)
\(=3^{1992}\times\left(9+3-1\right)\)
\(=3^{1992}\times11\)
\(\Rightarrow3^{1994}+3^{1993}-3^{1992}⋮11\)
d.
\(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\times\left(2^2+2-1\right)\)
\(=2^{24}\times\left(4+2-1\right)\)
\(=2^{24}\times5\)
\(\Rightarrow4^{13}+32^5-8^8⋮5\)
Chúc bạn học tốt
\(C=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\right)-\)\(\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+..+\frac{1}{2017}\)
\(\Rightarrow C=\left(\frac{1}{101}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\right)-\left(\frac{1}{1010}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)
\(\Rightarrow C=\frac{1}{2018}\)
\(1990\times1990-1992\times1998\)
\(=1990\times1990-\left(1992-2\right)\times\left(1998+2\right)\)
\(=1990\times1990-1990\times1990\)
\(=0\)
a) 1012 = (100+1)^2 = 100^2 + 200 + 1 = 10 000 + 200 +1 = 10 201
b) 1992 = (200 -1)^2 = 200^2 - 400 +1 = 40 000 - 400 +1 = 39 601
; c) 47.53 = (50-3) (50+3) = 50^2 - 3^2= 2500 -9 =2491
a) \(\left(100+1\right)^2=100^2+2\times100\times1+1^2=10000+200+1=10201\)
b)\(=\left(200-1\right)^2=200^2-2\times200\times1+1^2=40000-200+1=39801\)
c)\(=\left(50-3\right)\times\left(50+3\right)=50^2-3^2=2491\)