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Ta có công thức tổng quát như sau:
\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)
Áp dụng ta có:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
5,5 + \(\dfrac{3}{4}\) - 5 + \(\dfrac{1}{4}\)
(5,5 - 5) + (\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\))
= 0,5 + 1
= 1,5
\(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) : \(\dfrac{3}{2}\)
= \(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) x \(\dfrac{2}{3}\)
= (\(\dfrac{5}{2}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= (\(\dfrac{10}{4}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{4}\) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{6}\)
Tính \(x\):
435 - [\(x\) + 16] = 425 : 17
435 - [\(x\) + 16] = 25
[\(x\) + 16] = 435 - 25
\(x\) + 16 = 410
\(x\) = 410 - 16
\(x\) = 394
\(=5\cdot4\cdot\dfrac{9}{16}-4\cdot\left(-2\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}\cdot\left(-2\right)-\dfrac{9}{4}\)
\(=5\cdot\dfrac{9}{4}+4\cdot4\cdot\dfrac{3}{4}-1-\dfrac{9}{4}\)
\(=\dfrac{45}{4}-\dfrac{9}{4}+4\cdot3-1=9+12-1=20\)
5:
a: \(3^{2n}=\left(3^2\right)^n=9^n\)
\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)
=>\(3^{2n}>2^{3n}\)
b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
mà \(1568239201< 8036054027\)
nên \(199^{20}< 2003^{15}\)
4: \(100< 5^{2x-1}< 5^6\)
mà \(25< 100< 125\)
nên \(125< 5^{2x-1}< 5^6\)
=>3<2x-1<6
=>4<2x<7
=>2<x<7/2
mà x nguyên
nên x=3
