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\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
12/5 -4/5 =8/5
11/6 -2/3 =21/18 =7/6
7/8 -2/7 =33/56
4 -8/5 =20/5-8/5 =12/5
2 -3/8 =16/8 -3/8 =13/8
16/7 -2 =16/7 -14/7 =2/7
25/4 -3=25/4 -12/4 =13/4
câu 2 : câu 3 :
a) 8/5 a)4 - 8/5 = 20/5 - 8/5 =12/5
b) 7/6 b)2 - 3/8 =16/8 - 3/8 = 13/8
c) 35/56 c)16/7 - 2 = 16/7 - 14/7 = 2/7
d)25/4 - 3 = 25/4 - 12/4 = 13/4
\(a,\frac{3}{15}+\frac{2}{5}=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}.\)
\(b,\frac{9}{8}-\frac{5}{6}=\frac{7}{24}\)
\(c,\frac{1}{2}+\frac{3}{7}+\frac{11}{14}=\frac{13}{14}+\frac{11}{14}=\frac{24}{14}=\frac{12}{7}\)
\(d,\frac{8}{3}-\frac{1}{2}-1=\frac{13}{6}-1=\frac{7}{6}\)
~~~ học tốt ~~~
a) x+ \(\frac{4}{5}\)= \(\frac{4}{5}\)+ ( \(\frac{3}{7}\)+ \(\frac{3}{5}\))
x + \(\frac{4}{5}\)= \(\frac{4}{5}\)+ \(\frac{36}{35}\)
=> x = \(\frac{36}{35}\)
Vậy : x = \(\frac{36}{35}\)
b) \(\frac{4}{9}\)+\(\frac{8}{9}\)+ \(\frac{12}{9}\)+ ... + \(\frac{56}{9}\)
= \(\frac{4+8+12+...+56}{9}\)
Ta có : 4+8+12 + ... +56
Số các số hạng của dẫy số trên là : ( 56 -4) + 4 +1 = 14 ( số hạng )
=> Tổng trên = ( 4+56) x 14 : 2 = 420
=> \(\frac{4}{9}\)+ \(\frac{8}{9}\)+ ... + \(\frac{56}{9}\)= \(\frac{420}{9}\)= \(\frac{140}{3}\)
a)x=3/7+3/5=3x(1/7+1/5)=36/35
b)=(4+8+...+56)/9=(56+4)x14/9=280/3
Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?
A) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
2A-A = \(1-\dfrac{1}{32}\)
A= \(\dfrac{31}{32}\)
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)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)
\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)
c) \(\dfrac{5}{7}\) + \(\dfrac{7}{6}\) = \(\dfrac{30}{42}\) + \(\dfrac{49}{42}\) = \(\dfrac{79}{42}\)
\(\dfrac{7}{12}\) + \(\dfrac{17}{18}\) = \(\dfrac{21}{36}\) + \(\dfrac{34}{36}\) = \(\dfrac{55}{36}\)
\(\dfrac{9}{8}+\dfrac{15}{32}=\dfrac{36}{32}+\dfrac{15}{32}=\dfrac{51}{32}\)
\(4+\dfrac{35}{45}=\dfrac{36}{9}+\dfrac{7}{9}=\dfrac{43}{9}\)
d) \(\dfrac{11}{4}-\dfrac{15}{16}=\dfrac{44}{16}-\dfrac{15}{16}=\dfrac{29}{16}\)
\(\dfrac{5}{6}-\dfrac{5}{8}=\dfrac{40}{48}-\dfrac{30}{48}=\dfrac{10}{48}=\dfrac{5}{24}\)
\(\dfrac{196}{64}-2=\dfrac{49}{16}-\dfrac{32}{16}=\dfrac{17}{16}\)
\(3-\dfrac{13}{9}=\dfrac{27}{9}-\dfrac{13}{9}=\dfrac{14}{9}\)
e) \(\dfrac{8}{5}+\dfrac{7}{6}+\dfrac{5}{9}-2\)
= \(\dfrac{144}{90}+\dfrac{105}{90}+\dfrac{50}{90}-\dfrac{180}{90}\)
= \(\dfrac{249}{90}\) + \(\dfrac{50}{90}\) - \(\dfrac{180}{90}\)
= \(\dfrac{299}{90}\) - \(\dfrac{180}{90}\)
= \(\dfrac{119}{90}\)
3 - \(\dfrac{5}{6}\) - \(\dfrac{4}{9}\) + \(\dfrac{32}{24}\)
= 3 - \(\dfrac{5}{6}\) - \(\dfrac{4}{9}\) + \(\dfrac{4}{3}\)
= \(\dfrac{54}{18}\) - \(\dfrac{15}{18}\) - \(\dfrac{8}{18}\) + \(\dfrac{24}{18}\)
= \(\dfrac{39}{18}\) - \(\dfrac{8}{18}\) + \(\dfrac{24}{18}\)
= \(\dfrac{31}{18}\) + \(\dfrac{24}{18}\)
=\(\dfrac{55}{18}\)