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Bài 1:
\(\frac{1}{12},\frac{2}{3},\frac{3}{4},\frac{5}{6}\)
Bài 2:
A. \(\frac{11}{6}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{7}\)
D. \(\frac{15}{8}\)
Tích đúng cho mình nhé!
a)2/5+3/10-1/2=7/10-1/2=2/10=1/5
b)8/11+8/33x3/4=8/11+2/11=10/11
c)7/9x3/14:5/8=1/6:5/8=8/30+4/15
d)5/12-7/32:21/16=5/12-1/6=3/12
nha
a) 2/5+3/10-1/2
= 4/10+3/10-5/10
= 2/10 = 1/5
b) 8/11+8/33x3/4
= 8/11+2/11
= 10/11
c) 7/9x3/14:5/8
= 1/6:5/8
= 1/6x8/5
=4/15
d) 5/12-7/32:21/16
= 5/12-7/32x16/21
= 5/12-1/6
= 5/12-2/12
= 3/12
1.\(\dfrac{8}{11}+\dfrac{8}{33}\times\dfrac{3}{4}=\dfrac{8}{11}+\dfrac{8\times3}{33\times4}=\dfrac{8\times12}{11\times12}+\dfrac{8\times3}{33\times4}=\dfrac{8\times15}{132}=\dfrac{120}{132}=\dfrac{10}{11}\)
2.\(\dfrac{7}{9}\times\dfrac{3}{14}:\dfrac{5}{8}=\dfrac{7}{9}\times\dfrac{3}{14}\times\dfrac{8}{5}=\dfrac{4}{15}\)
3.\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}=\dfrac{5}{12}-\dfrac{7}{32}\times\dfrac{16}{21}=\dfrac{5}{12}-\dfrac{1}{6}=\dfrac{5-2}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
`@V.Tr.V`
\(a)\frac{2}{5}+\frac{3}{10}-\frac{1}{2}\)
\(=\frac{4}{10}+\frac{3}{10}-\frac{5}{10}=\frac{1}{5}\)
\(b)\frac{8}{11}+\frac{8}{33}.\frac{3}{4}\)
\(=\frac{8}{11}+\frac{2}{11}=\frac{10}{11}\)
\(c)\frac{7}{9}.\frac{3}{14}:\frac{5}{8}\)
\(=\frac{1}{6}:\frac{5}{8}=\frac{4}{15}\)
\(d)\frac{5}{12}-\frac{7}{32}:\frac{21}{16}\)
\(=\frac{5}{12}-\frac{1}{6}=\frac{5}{12}-\frac{2}{12}=\frac{1}{4}\)
2/5 + 3/10 - 1/2
= 4/10 + 3/10 -5/10
=2/10
=1/5.
8/11 + 8/33 *3/4
=8/11 + 2/11
=10/11
7/9 *3/14 : 5/8
=1/3 *1/2 : 5/8
=1/6: 5/8
=1/3 * 4/5
=4/15
5/12 - 7/32 : 21/16
=5/12 - 1/6
=5/12 - 2/12
=1/4
2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5
#)Giải :
Đặt \(A=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+\frac{7}{32}+...+\frac{7}{512}\)
\(A=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\)
\(\Rightarrow2A=\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\)
\(\Rightarrow2A-A=\left(\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+\frac{7}{2^5}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+\frac{7}{2^4}+...+\frac{7}{2^9}\right)\)
\(\Rightarrow A=\frac{7}{2^{10}}-\frac{7}{2}\)
tiếc nhỉ giải theo kiểu học sinh lớp 4 cơ