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a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
cái đầu =\(\frac{127}{128}\)vì:
1/2+1/4=3/4 mà 3/4 =1-1/4
1/2+1/4+1/8=7/8 mà 7/8=1-1/8
ta suy ra cách làm: Tổng dãy phân số trên bằng 1 trừ cho phân số cuối
=> Tổng dãy trên =1-1/128= 127/128
\(\left(a\right)\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{36+15}{32}=\frac{51}{32}\)
\(\left(b\right)4+\frac{35}{45}=\frac{4}{1}+\frac{35}{45}=\frac{180}{45}+\frac{35}{45}=\frac{180+35}{45}=\frac{43}{9}\)
\(\left(c\right)\frac{11}{4}-\frac{15}{16}=\frac{44}{16}-\frac{15}{16}=\frac{44-15}{16}=\frac{29}{16}\)
\(\left(d\right)3-\frac{13}{9}=\frac{3}{1}-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{27-13}{9}=\frac{14}{9}\)
\(\left(e\right)\frac{5}{6}-\frac{5}{8}=\frac{20}{24}-\frac{15}{24}=\frac{20-15}{24}=\frac{5}{24}\)
\(\left(g\right)\frac{196}{64}-2=\frac{196}{64}-\frac{2}{1}=\frac{196}{64}-\frac{128}{64}=\frac{196-128}{64}=\frac{17}{16}\)
Mình đảm bảo đúng ! Chúc bạn học tốt!
a) \(\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{51}{32}\)
b) \(4+\frac{35}{45}=4+\frac{7}{9}=\frac{36}{9}+\frac{7}{9}=\frac{43}{9}\)
c)\(\frac{11}{4}+\frac{15}{16}=\frac{44}{16}+\frac{15}{16}=\frac{59}{16}\)
d )\(3-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{14}{9}\)
e ) \(\frac{5}{6}+\frac{5}{8}=\frac{40}{48}+\frac{30}{48}\)\(=\frac{70}{48}\)
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
Bài 1:
a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)
\(\frac{2\cdot x-4,36}{0,125}=8\)
\(2\cdot x-4,36=8\cdot0,125\)
\(2\cdot x-4,36=1\)
\(2\cdot x=1+4,36\)
\(2\cdot x=5,36\)
\(x=\frac{5,36}{2}=2,68\)
b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)
\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)
\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)
Bài 2:
a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )
\(x+5,2=4,7\cdot3,2+0,5\)
\(x+5,2=15,54\)
\(x=15,54-5,2=10,34\)
b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)
Bài 3:
a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)
\(x\cdot\left(104,5-14,1+9,6\right)=25\)
\(x\cdot100=25\)
\(x=\frac{25}{100}=\frac{1}{4}=0,25\)
b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
1/1+2 + 1/1+2+3 +1/1+2+3+4 +...+1/1+2+3+...+50
Ta có 2/2(1+2)+2/2(1+2+3)+...+2/2(1+2+...+50)
=2/6+2/12+2/20+...+2/2550
=2/2.3+2/3.4+...+2/50.51
=2(1/2.3+1/3.4+...+1/50.51)
=2(1/1-1/2+1/2-...+1/50-1/51)
=2.(1-1/51)
=2.50/51=100/51
(-3,2) . -15/64 + (0,8 - 34/15) : 7/2
=(-3,2) . -15/64 + -22/15 :7/2
=3/4 + -44/105
=139/420
**** mik nha