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\(M\times88=88\times1090\\ M=1090\\ \left(N+1090\right)\times256=\left(984+1090\right)\times256\\ N=984\)
Ta có:
\(M=2007\times2007=\left(2004+3\right)\times2007=2004\times2007+3\times2007\)
\(N=2004\times2008=2004\times\left(2007+1\right)=2004\times2007+2004\times1\)
\(3\times2007=6021;2004\times1=2004\)
Vì \(2004\times2007=2004\times2007\)và\(6021>2004\)nên \(M>N\).
m = 2007 x 2007 = (2004 + 3) x 2007 = 2004 x 2007 + 3 x 2007 = 2004 x 2007 + 6021
n = 2004 x 2008 = 2004 x (2007 + 1) = 2004 x 2007 + 2004 x 1 = 2004 x 2007 + 2004
Ta có : 2004 x 2007 + 6021 \(>\) 2004 x 2007 + 2004
Nên, m \(>\) n.
a) 1763/1765>191/193;
171/172<1090/1091;
576/579<195/196
b) 1385/1386>176/177;
185/187<1070/1071
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)