1+4=5
2+5=12
3+6=21
8+11=?
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1 + 4 = 5
2 + 5 = 12 ( 2 + 5 + 5 = 12)
3 + 6 = 21 (3 + 6 + 12 = 21)
8 + 11 = 40 ( 8 + 11 + 21= 40)
Quy luật Cộng thêm kết quả của dãy số ở trên
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: \(=-62-38+1523-2523-92\)
\(=-100+1000-92=900-92=808\)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a: =6/35x39/54=13/105
b: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}=\dfrac{-5}{7}+\dfrac{12}{7}=1\)
1) \(\left(+15\right)+\left(+17\right)=15+17=32\)
2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)
3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)
4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)
5) \(\left(-15\right)+20=20-15=5\)
6) \(\left(-5\right)+8+7+5\)
\(=\left(-5+5\right)+\left(8+7\right)\)
\(=15\)
7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)
\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)
\(=\left(-10\right)+\left(-11\right)\)
\(=-21\)
8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)
\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)
\(=10+\left(-30\right)\)
\(=-20\)
9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)
\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)
\(=\left(-17\right)+\left(-100\right)\)
\(=-117\)
10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)
\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)
\(=\left(-400\right)+40+240\)
\(=\left(-360\right)+240\)
\(=-120\)
11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)
\(=\left(-596\right)+2001+1999-404+189\)
\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)
\(=\left(-1000\right)+2190+1999\)
\(=1190+1999\)
\(=3189\)
12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
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