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g: \(=-457+237+23-123=-220-100=-320\)
h: \(=\left(1-3\right)+\left(5-7\right)+...+\left(41-43\right)+\left(45-47\right)\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+\left(-2\right)\)
\(=-2\cdot12=-24\)
i: \(=173+27-46-54-19=200-100-19=100-19=81\)
k: \(=-52+82+49-15+13-36\)
\(=30+34-23\)
=30+11
=41
l: \(=\left(3-5\right)+\left(7-9\right)+\left(11-13\right)+\left(15-17\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)
=-8
m: \(=\left(1-2\right)+\left(3-4\right)+...+\left(2001-2002\right)+2003\)
\(=2003-1-1-...-1\)
\(=2003-1001=1002\)
n:Số số hạng là:
\(\left[\left(-51\right)-\left(-99\right)\right]:1+1=49\left(số\right)\)
Tổng là \(\left(-51-99\right)\cdot\dfrac{49}{2}=-3675\)
o: \(=-62-38+1523-2523-92\)
\(=-100+1000-92=900-92=808\)
a: \(\dfrac{6}{7}:\left(\dfrac{2}{5}\cdot\dfrac{6}{7}\right)\)
\(=\dfrac{6}{7}:\dfrac{12}{35}\)
\(=\dfrac{6}{7}\cdot\dfrac{35}{12}=\dfrac{6}{12}\cdot\dfrac{35}{7}=\dfrac{5}{2}\)
b: \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)
\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)
\(=1-\dfrac{8}{9}=\dfrac{1}{9}\)
c: \(\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\)
\(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\)
\(=\dfrac{48+7-42}{56}=\dfrac{13}{56}\)
d: \(\dfrac{-1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{4}{5}\)
\(=-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5-15+24}{30}=\dfrac{4}{30}=\dfrac{2}{15}\)
a: 14/5-7/5=7/5
b: 7/8-1/3+5/4
=21/24-8/24+30/24
=43/24
c; =7/6+5/6+2/15+13/15
=2+1
=3
d: =4*5/3*11=20/33
e: =2/9*1/6*1/4=2/9*1/24=1/108
2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)
b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)
c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)
3: \(=20-12-8+12=20-8=12\)
5: \(=-18-42-21-35=-116\)
3: \(=-15+18-12+8=-27+26=-1\)
2: \(=-12+21-15+10=9-5=4\)
2) -3(4 - 7) + 5(-3 + 2)
= -3.4 + 3.7 - 5.3 + 5.2
= -12 + 21 -15 + 10
= 31 - 27
= 4
4) -5(2 - 7) + 4(2 - 5)
= -5.2 + 5.7 + 4.2 - 4.5
= -10 + 35 + 8 - 20
= 38 - 30
= 8
** Bổ sung điều kiện $n$ là số tự nhiên.
Lời giải:
Hiển nhiên $2003^n$ luôn lẻ với mọi số tự nhiên $n$
$\Rightarrow 2003^n+5\vdots 2$
$\Rightarrow (2003^n+5)(2003^n+7)\vdots 2(1)$
Lại có:
Nếu $n$ lẻ:
$2003\equiv -1\pmod 3\Rightarrow 2003^n+7\equiv (-1)^n+7\equiv -1+7\equiv 0\pmod 3$
Nếu $n$ chẵn:
$2003\equiv -1\pmod 3\Rightarrow 2003^n+5\equiv (-1)^n+5\equiv 1+5\equiv 0\pmod 3$
Vậy $(2003^n+5)(2003^n+7)\vdots 3(2)$
Từ $(1); (2)$ mà $(2,3)=1$ nên $(2003^n+5)(2003^n+7)\vdots (2.3=6)$