1, SO SÁNH :
A= 1930+ 5/ 1931+ 5
B= 1931+ 5 / 1932+ 5
2, TÍNH:
C= 1+ 1/2 ( 1+ 2 )+ 1/3 ( 1+ 2+ 3 )+ 1/4 ( 1+ 2+ 3+ 4 )+...+ 1/200( 1+ 2+ 3+...+ 200 )
( LƯU Ý: Dấu ''/'' là gạch của phân số, vd: a/b là a phần b )
PLEASE HELP ME
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a: =3/8-1/4
=3/8-2/8
=1/8
b: =-5/9+3/5-1/9+2/5
=-2/3+1
=1/3
c: =21/7*5/25=3/5
d: =3/4+11/10:(2/5-3/2)-1/9
=-13/36
1)Ta thấy nếu số đó công với 4 thì chia hết cho cả 3 số
Gọi số phải tìm là A
Ta có A + 4 chia hết cho 5 , 7 , 9
Mà A nhỏ nhất nên A + 4 = 5 . 7 . 9 = 315
Do đó A = 315 - 4 = 311
2)a)Ta có S = 2^1 + 2^2 +2^3 +...+ 2^100
S = ( 2^1 + 2^2 + 2^3 +2^4 ) +...+( 2^97 + 2^98 + 2^99 + 2^100 )
S = 1( 2^1 + 2^2 + 2^3 + 2^4 ) +...+ 2^96( 2^1 + 2^2 + 2^3 + 2^4 )
S = 1.30 +...+2^96.30
S = ( 1 +...+2^96 )30
Vì 30 chia hết cho 15 nên ( 1 +...+2^96 )30 chia hết cho 15
Hay S chia hết cho 15
b) Vì S cha hết cho 30 nên S chia hết cho 10
Suy ra S có tận cùng là 0
c) S = 2^1 + 2^2 + 2^3 +...+2^100
2S = 2^2 + 2^3 + 2^4 +...+ 2^101
2S - S =( 2^2 + 2^3 +...+ 2^101 ) - ( 2^1 + 2^2 + ... + 2^100 )
S = 2^101 - 2^1
S = 2^101 - 2
1. 158
2a. 0 ( doan nha )
b.S = ( 2 + 2^2 +2^3+2^4) + ( 2^5 + 2^6 + 2^7 + 2^8 ) +...+ ( 2^97 + 2^ 98 + 2^99 +2^100 )
= 2.( 1+2+2^2+2^3 ) + 2^5. ( 1+2+2^2+2^3)+2^97.( 1+2+2^2+2^3)
= 2.15+2^5.15+...+2^97.15
= 15.(2+2^5+...+2^97) chia het 15
c.2^101-2^1
3. chiu !
Bài 1: Tính
\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)
\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)
\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)
\(=\dfrac{5}{8}.\dfrac{-4}{15}\)
\(=\dfrac{-1}{6}\)
\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{3}{4}\)
\(=\dfrac{-51}{40}\)
\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)
\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)
\(=\dfrac{4}{6}\)
\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)
\(=\dfrac{4}{3}-1\)
\(=\dfrac{1}{3}\)
\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)
\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)
\(=1:\dfrac{1}{5}\)
\(=5\)
\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)
\(=3+\dfrac{11}{7}\)
\(=3\dfrac{11}{7}=\dfrac{32}{7}\)
\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)
\(=1:\dfrac{19}{5}\)
\(=\dfrac{5}{19}\)
\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)
\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+1\right)\)
\(=1:\dfrac{5}{3}\)
\(=\dfrac{3}{5}\)
\(\text{9)}\)
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)
\(=\dfrac{330875}{1507764}\)
1, Vì A, B < 1
\(\Rightarrow B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)
2, Đề là thế này?? \(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)
\(\Rightarrow C=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(\Rightarrow C=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(\Rightarrow C=\frac{\left(2+201\right).200}{4}=10150\)