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a) \(4^{13}+4^{14}+4^{15}+4^{16}=4^{13}\left(1+4\right)+4^{14}\left(1+4\right)=4^{13}.5+4^{14}.5=5\left(4^{13}+4^{14}\right)⋮5\Rightarrow dpcm\)
c) \(2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}\)
\(=2^{10}\left(1+2+2^2\right)+2^{13}\left(1+2+2^2\right)\)
\(=2^{10}.7+2^{13}.7=7\left(2^{10}+2^{13}\right)⋮7\Rightarrow dpcm\)
Câu c bạn xem lại đê
\(A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)
\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(A=\frac{1}{3}-\frac{1}{111}\)
\(A=\frac{12}{37}\)
mà dài quá bạn ơi ban tách ra thành nhiều câu hỏi đi thế này trả lời lâu lắm
a: =-5-7-8-3=-26
b: =-4-2-5-6=-17
c: =-8-11-4-2=-25
d: -9-15-6-3=-30
e: =-12-9-3-8=-32
\(C=\left(1-2-3-4\right)+...+\left(197-198-199-200\right)\)
=-8x25=-200
\(D=-\left(11+13+...+99\right)+\left(10+12+...+98\right)\)
=(-1)+(-1)+...+(-1)
=-1x45=-45
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
1 + 1 = 2
2 + 2 = 4
3 + 3 = 6
4 + 4 = 8
5 + 5 = 10
6 + 6 = 12
7 + 7 = 14
8 + 8 = 16
9 + 9 = 18
10 + 10 = 20
11 + 11 = 22
12 + 12 = 24
13 + 13 = 26
14 + 14 = 28
15 + 15 = 30
(1+2-3-4+5-6-7 )+(8+9-10-11+12)-(13-14+15)=?
= -2 + 8 - 14
=6 - 14
= - 8
hok tốt
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) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
\(\dfrac{10^2+11^2+12^2}{15^2+14^2}\)
\(=\dfrac{100+121+144}{225+196}\)
\(=\dfrac{221+144}{421}\)
\(=\dfrac{365}{421}\)