(1+2+3+4+...+9)x (21x5-21-4x21)
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\(\text{(1+2+3+4+....+9)x(21x5-21-4x21}\))
\(=\text{(1+2+3+4+....+9)x}\) \(\left(105-21-84\right)\)
\(=\left(1+2+3+4+....+9\right)\times0\)
\(=0\)
\(#Siliver\)
Ta có: \(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{x+3}{x+3}-\dfrac{1}{x+3}\right)\)
\(=\dfrac{21+x^2+3x-4x-12-\left(x^2-4x+3\right)}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{x^2-x+9-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+2}{x+3}\)
\(=\dfrac{3x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+2}{x+3}\)
\(=\dfrac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}\)
\(=\dfrac{3}{x-3}\)
có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
\(\dfrac{8}{9}+\dfrac{3}{18}=\dfrac{16}{18}+\dfrac{3}{18}=\dfrac{19}{18}\) \(\dfrac{5}{7}-\dfrac{4}{11}=\dfrac{55}{77}-\dfrac{28}{77}=\dfrac{3}{11}\)
\(\dfrac{3}{4}+\dfrac{4}{5}=\dfrac{15}{20}+\dfrac{16}{20}=\dfrac{21}{20}\) \(\dfrac{5}{7}-\dfrac{3}{5}=\dfrac{25}{35}-\dfrac{21}{35}=\dfrac{4}{35}\)
\(\dfrac{7}{9}X\dfrac{3}{4}=\dfrac{7}{9}X\dfrac{3}{4}=\dfrac{7}{12}\) \(\dfrac{8}{3}X\dfrac{21}{4}=\dfrac{8}{3}X\dfrac{21}{4}=14\)
\(\dfrac{12}{7}:\dfrac{1}{4}=\dfrac{12}{7}X\dfrac{4}{1}=\dfrac{48}{7}\) \(\dfrac{7}{2}:\dfrac{9}{4}=\dfrac{7}{2}X\dfrac{4}{9}=\dfrac{14}{9}\)
ta có : ( 1 + 2 + 3 + 4 + ..... + 9 ) . ( 21 . 5 - 21 - 4 . 21 )
= ( 1 + 2 + 3 + 4 + .....+ 9 ) . [ 21 ( 5 - 1 - 4 ) ]
= ( 1 + 2 + 3 + 4 + ..... + 9 ) . ( 21 . 0 )
= ( 1 + 2 + 3 + 4 + ...... + 9 ) . 0
= 0
[ dấu " . " là dấu nhân á nhaaa ]