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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
Nay rảnh lại qua làm mấy bài lớp 4 ^^
Tui làm luôn kh ghi đề nhá
1,
\(\frac{2}{3}+\frac{8}{15}=\frac{6}{5}\)
\(\frac{2}{5}+\frac{2}{3}=\frac{16}{15}\)
\(\frac{2}{3}+\frac{5}{6}=\frac{3}{2}\)
2,
\(\cdot\frac{3}{7}+\frac{6}{5}+\frac{4}{7}\)
\(=\frac{3}{7}+\frac{4}{7}+\frac{6}{5}\)
\(=1+\frac{6}{5}\)
\(=\frac{11}{5}\)
\(\cdot\frac{6}{4}+\frac{7}{3}+\frac{4}{4}+\frac{3}{3}\)
\(=\frac{6}{4}+\frac{4}{4}+\frac{7}{3}+\frac{3}{3}\)
\(=\frac{10}{4}+\frac{10}{3}\)
\(=\frac{35}{6}\)
23+815=6523+815=65
25+23=161525+23=1615
23+56=3223+56=32
2,
⋅37+65+47⋅37+65+47
=37+47+65=37+47+65
=1+65=1+65
=115=115
⋅64+73+44+33⋅64+73+44+33
=64+44+73+33=64+44+73+33
=104+103=104+103
=356
\(\frac{7}{15}+\frac{9}{16}+\frac{8}{15}+\frac{7}{16}\)
\(=\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{9}{16}+\frac{7}{16}\right)\)
\(=\frac{15}{15}+\frac{16}{16}\)
\(=1+1\)
\(=2\)
\(a)\frac{2}{5}+\frac{3}{10}-\frac{1}{2}\)
\(=\frac{4}{10}+\frac{3}{10}-\frac{5}{10}=\frac{1}{5}\)
\(b)\frac{8}{11}+\frac{8}{33}.\frac{3}{4}\)
\(=\frac{8}{11}+\frac{2}{11}=\frac{10}{11}\)
\(c)\frac{7}{9}.\frac{3}{14}:\frac{5}{8}\)
\(=\frac{1}{6}:\frac{5}{8}=\frac{4}{15}\)
\(d)\frac{5}{12}-\frac{7}{32}:\frac{21}{16}\)
\(=\frac{5}{12}-\frac{1}{6}=\frac{5}{12}-\frac{2}{12}=\frac{1}{4}\)
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
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
\(=\left(\frac{4}{17}+\frac{13}{17}\right)-\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{10}{16}-\frac{7}{16}\right)\)
\(=1-1+\frac{3}{16}\)
\(\frac{3}{16}\)
gộp 1,3
2,4
5,6
thì ra 17/17-15/15+3/16=3/16
hk tốt!
a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)
b)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)
\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)
#)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
#)Nếu là cách lớp 4 thì mình chịu @@
hazzz
\(giải:\)
Đặt \(A\)\(=\frac{7}{2}\)\(+\)\(\frac{7}{4}\)\(+\)\(\frac{7}{8}\)\(+\)\(\frac{7}{16}\)\(+\)\(\frac{7}{32}\)\(+\)\(...\)\(+\)\(\frac{7}{512}\)
\(A\)\(=\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)
\(\Rightarrow\)\(2A\)\(-\)\(A\)\(=\)\((\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(\frac{7}{2^5}\)\(+\)\(...\)\(+\)\(\frac{7}{2^{10}}\)\()\)\(-\)
\((\)\(\frac{7}{2}\)\(+\)\(\frac{7}{2^2}\)\(+\)\(\frac{7}{2^3}\)\(+\)\(\frac{7}{2^4}\)\(+\)\(...\)\(+\)\(\frac{7}{2^9}\)\()\)
\(\Rightarrow\)\(A=\)\(\frac{7}{2^{10}}\)\(-\)\(\frac{7}{2}\)
\(P=\frac{7}{2}+\frac{7}{4}+\frac{7}{8}+\frac{7}{16}+...+\frac{7}{512}\)
\(=\frac{7}{2}+\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^9}\)
\(\Rightarrow2P-P=\left(\frac{7}{2^2}+\frac{7}{2^3}+...+\frac{7}{2^{10}}\right)-\left(\frac{7}{2}+\frac{7}{2^2}+...+\frac{7}{2^9}\right)\)
\(\Rightarrow P=\frac{7}{2^{10}}-\frac{7}{2}\)