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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
a: 2/9=4/18
1/3=6/18
5/18=5/18
b: 7/15=14/30
1/5=6/30
-5/6=-25/30
c: -21/56=-3/7
-3/16=-63/336
5/24=70/336
-21/56=-3/7=-144/336
d: \(\dfrac{-4}{7}=\dfrac{-36}{63}\)
8/9=56/63
\(-\dfrac{10}{21}=-\dfrac{30}{63}\)
e: 3/-20=-3/20=-9/60
-11/-30=11/30=22/60
7/15=28/60
a) 45 ⋮ x
Vì 45 ⋮ x nên x E Ư( 45 )
= { 1;3;5;9;15;45 }
mà x E Ư(45)
=> x E { 1;3;5;9;15;45 }
b) 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x và x lớn nhất
Vì 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x nên x E ƯC ( 24;36;160)
mà x lớn nhất
=> x E ƯCLN ( 24;36;160 )
Ta có
24 = 23 . 3
36 = 22.32
160 = 25 . 5
=> ƯCLN ( 24;36;160 ) = 22 = 4
\(A=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+...-\left(\frac{1}{8}+\frac{1}{9}\right)\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(S=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(S=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
\(S=3\left(\frac{1}{1.2}\right)-5\left(\frac{1}{2.3}\right)+7\left(\frac{1}{3.4}\right)-9\left(\frac{1}{4.5}\right)+11\left(\frac{1}{5.6}\right)-13\left(\frac{1}{6.7}\right)+15\left(\frac{1}{7.8}\right)-17\left(\frac{1}{8.9}\right)\)
\(S=3\left(1-\frac{1}{2}\right)-5\left(\frac{1}{2}-\frac{1}{3}\right)+7\left(\frac{1}{3}-\frac{1}{4}\right)-9\left(\frac{1}{4}-\frac{1}{5}\right)+11\left(\frac{1}{5}-\frac{1}{6}\right)-13\left(\frac{1}{6}-\frac{1}{7}\right)+15\left(\frac{1}{7}-\frac{1}{8}\right)-17\left(\frac{1}{8}-\frac{1}{9}\right)\)
\(S=\left(3-\frac{3}{2}\right)-\left(\frac{5}{2}-\frac{5}{3}\right)+\left(\frac{7}{3}-\frac{7}{4}\right)-\left(\frac{9}{4}-\frac{9}{5}\right)+\left(\frac{11}{5}-\frac{11}{6}\right)-\left(\frac{13}{6}-\frac{13}{7}\right)+\left(\frac{15}{7}-\frac{15}{8}\right)-\left(\frac{17}{8}-\frac{17}{9}\right)\)\(S=3-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+\frac{11}{5}-\frac{11}{6}-\frac{13}{6}+\frac{13}{7}+\frac{15}{7}-\frac{15}{8}-\frac{17}{8}+\frac{17}{9}\) Giờ bạn chỉ cần nhóm từng cặp phân số có cùng tử số rồi tính tiếp là ra kết quả thôi
( khi nhóm cặp nhớ đổi dấu nha)
\(=\dfrac{15}{7\cdot8}-\dfrac{13}{6\cdot7}+\dfrac{11}{5\cdot6}-\dfrac{9}{4\cdot5}+\dfrac{7}{3\cdot4}-\dfrac{5}{2\cdot3}+\dfrac{3}{1\cdot2}\)
\(=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{2}\)
=1+1/8=9/8
