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\(A=2015\cdot2015=2015^2\)
\(B=2016\cdot2014=\left(2015+1\right)\left(2015-1\right)=2015^2-1\)
=>\(A>B\)
**** cho minn nhe
637.527-189 =637.526+637.1-189 =637.526+448 =1
526.637+448 637.526+448 637.526+448
\(a)\)\(\left(1989.1990+3978\right)\div\left(1992.1991-3984\right)\)
\(=\)\(\left(1989.1990+1989.2\right)\div\left(1992.1991-1992.2\right)\)
\(=\)\(\left[1989.\left(1990+2\right)\right]\div\left[1992.\left(1991-2\right)\right]\)
\(=\)\(\left(1989.1991\right)\div\left(1992.1989\right)\)
\(=\)\(1\)
\(b)\)\(\left(637.527-189\right)\div\left(526.637+448\right)\)
\(=\)\(\left[637.\left(526-1\right)-189\right]\div\left(526.637-448\right)\)
\(=\)\(\left[637.526+\left(637-198\right)\right]\div\left(637.526-448\right)\)
\(=\)\(\left(636.526+448\right)\div\left(637.526-448\right)\)
\(=\)\(1\)
(637.527-189):(526.637+448)
= \(\frac{637.527-189}{526.637+448}\)
= \(\frac{637.\left(526+1\right)-189}{526.637+448}\)
= \(\frac{637.526+637-189}{526.637+448}\)
= \(\frac{637.526+448}{526.637+448}\)
= \(1\)
Ta có:
\(\dfrac{637.527-189}{526.637+448}=\dfrac{637.\left(526+1\right)-189}{526.637+448}=\dfrac{637.526+637-189}{526.637+448}=\dfrac{637.526+448}{526.637+448}=1\)
Ta có: B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<1 ( Vì 172009+1< 172010+1 )
Nên B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<\(\frac{17^{2009}+1+16}{17^{2010}+1+16}\)
=\(\frac{17^{2009}+17}{17^{2010}+17}\)
=\(\frac{17\left(17^{2008}+1\right)}{17\left(17^{2009}+1\right)}\)
=\(\frac{17^{2008+1}}{17^{2009}+1}\)=A
Vậy A>B
\(B=3^2+3^3+...+3^{99}\)
\(3B=3^3+3^4+...+3^{100}\)
\(3B-B=\left(3^3+3^4+...+3^{100}\right)-\left(3^2+3^3+...+3^{99}\right)\)
\(2B=3^{100}-3^2\)
\(B=\frac{3^{100}-9}{2}\)
\(2B+9=3^{2n+4}\)
\(\Leftrightarrow3^{2n+4}=3^{100}\)
\(\Leftrightarrow2n+4=100\)
\(\Leftrightarrow n=48\).
a) \(637.527-189\) và \(526.637+448\)
Ta có: \(637.526+448=526.637+448\)
\(\Rightarrow637.527-189=526.637+448\)
b) \(2016.2014\) và \(2015.2015\)
Ta có: \(\hept{\begin{cases}2016.2014=2015.2014+2014\\2015.2015=2015.2014+2015\end{cases}}\)
Mà: \(2015.2014=2015.2014\) và \(2015>2014\)
\(\Rightarrow2015.2014+2014< 2015.2014+2015\)
\(\Rightarrow2016.2014< 2015.2015\)