a) x +\(\frac{4}{5}\)= \(\frac{4}{5}+\left(\frac{3}{7}+\frac{3}{5}\right)\) b) \(\frac{4}{9}+\frac{8}{9}+\frac{12}{9}+\frac{16}{9}+...+\frac{48}{9}+\frac{52}{9}+\frac{56}{9}\)
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a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
\(=\frac{5}{5}+\frac{7}{7}=1+1=2\)
a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
= 1 + 1
= 2
b) \(\frac{4}{9}+\frac{8}{9}+\frac{12}{9}+\frac{16}{9}+...+\frac{48}{9}+\frac{52}{9}+\frac{56}{9}\)
\(=\frac{4+8+12+16+...+48+52+56}{9}\)
Xét 4 + 8 + 12 + 16 + ... + 48 + 52 + 56
Số các số hạng là:
(56 - 4) : 4 + 1 = 14 (số)
4 + 8 + 12 + 16 + ... + 48 + 52 + 56 = (56 + 4) x 14 : 2 = 420
\(\frac{4+8+12+16+...+48+52+56}{9}=\frac{420}{9}=\frac{140}{3}\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
\(A=\left(-\frac{5}{11}\right).\frac{7}{15}+\frac{11}{-5}.\frac{30}{33}\)
\(A=-\frac{7}{33}+-2\)
\(A=-\frac{73}{33}\)
[ A] = -2
a. \(1\frac{5}{7}\)-\(\frac{9}{7}\)*\(\frac{16}{9}\)
=\(\frac{12}{7}\)-\(\frac{16}{7}\)
=\(\frac{-4}{7}\)
b. \(\frac{-5}{8}\):\(\frac{1}{4}\)-\(\frac{6}{13}\)*4+\(\frac{3}{8}\)
=\(\frac{-5}{8}\cdot\)4-\(\frac{6}{13}\)*4+\(\frac{3}{8}\)
=4*(\(\frac{-5}{8}\)-\(\frac{6}{13}\))+\(\frac{3}{8}\)
=4*\(\frac{-113}{104}\)+\(\frac{3}{8}\)
=\(\frac{-113}{26}\)+\(\frac{3}{8}\)
=\(\frac{-413}{104}\)
c.( \(\frac{3}{8}\)+\(\frac{-1}{4}\)-\(\frac{5}{12}\)):\(\frac{1}{3}\)
=\(\frac{-7}{24}\)*3
=\(\frac{-7}{8}\)
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