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\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\Leftrightarrow\frac{4}{20}-x=\frac{21}{128}.\frac{32}{9}\)
\(\Leftrightarrow\frac{1}{5}-x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{1}{5}-\frac{7}{12}\)
\(\Leftrightarrow x=-\frac{23}{60}\)
Vậy \(x=\frac{-23}{60}\)
Bài làm
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{3}{20}+\frac{1}{20}-x\right)=\frac{21}{128}.\frac{32}{9}\)
\(\frac{4}{20}-x=\frac{7}{4}.\frac{1}{3}\)
\(\frac{4}{20}-x=\frac{7}{12}\)
\(x=\frac{4}{20}-\frac{7}{12}\)
\(x=\frac{12}{60}-\frac{35}{60}\)
\(x=-\frac{23}{60}\)
Vậy \(x=-\frac{23}{60}\)
\(x=-\frac{37}{15}\)
a) \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Leftrightarrow\)\(A=2-\frac{1}{2^7}=\frac{255}{128}\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{2}{7}=\frac{1}{7}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
a, 1727 + [ 6993 : 111 + ( 848 - 95 ) ] x 4 - 2
= 1727 + [ 63 + 753 ] x 4 - 2
= 1727 + 816 x 4 - 2
= 1727 + 3264 - 2
= 4991 - 2
= 4989
b, 75/100 + 18/21 + 19/32 + 1/4 + 3/21 + 13/32
= 75/100 + 1/4 + ( 18/21 + 3/21 ) + ( 19/32 + 13/32 )
= 75/100 + 25/100 + 21/21 + 32/32
= 100/100 + 1 + 1
= 1 + 1 + 1 = 3
c, 4 2/5 + 5 6/9 + 2 3/4 + 3/5 + 1/3 + 1/4
= ( 4 2/5 + 3/5 ) + ( 2 3/4 + 1/4 ) + 5 6/9 + 1/3
= 5 + 3 + 5 6/9 + 3/9
= 5 + 3 + 6
= 8 + 6 = 14
d, 3/4 + 25/36 - ( 4/9 + 13/18 + 1/72 )
= 27/36 + 25/36 - ( 32/72 + 52/72 + 1/72 )
= 52/36 - ( 84/72 + 1/72 )
= 52/36 - 85/72
= 104/72 - 85/72
= 19/72
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
bài 1 tính nhanh
mik xin sửa đề câu a thành thế này ~
\(a,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2-A=\) ( \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) ) - ( \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(B\cdot3-B=\) ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) )
\(B\cdot2=\) \(1-\frac{1}{729}\)
\(B\cdot2=\frac{728}{729}\)
\(B=\frac{728}{729}:2\)
\(B=\frac{364}{729}\)
\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=\frac{1}{1}-\frac{1}{6}\)
\(C=\frac{5}{6}\)
\(\left(\frac{3}{20}+\frac{1}{2}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{13}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\frac{13}{20}-x=\frac{21}{128}\times\frac{32}{9}\)
\(\frac{13}{20}-x=\frac{7}{12}\)
\(x=\frac{13}{20}-\frac{7}{12}\)
\(x=\frac{1}{15}\)
( 3/20 + 1/2 - x ) : 32/9 = 21/128
3/20+ 1/2 - x = 21/128 x 32/9
13/20 - x = 7/12
x = 13/20 - 7/12
x = 1/15