bài 4: tính bằng cách thuận tiện nhất:
a) 7/15 + 14/15+ 13/15
b) 13/4 + 4/9 + 7/8
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\(\frac{7}{4}\times\frac{15}{7}+\frac{13}{7}\times\frac{9}{4}-\frac{2}{7}\)
\(=\left(\frac{7}{4}+\frac{9}{4}\right)\times\left(\frac{15}{7}+\frac{13}{7}\right)-\frac{2}{7}\)
\(=4\times4-\frac{2}{7}\)
\(=16-\frac{2}{7}\)
\(=\frac{112}{7}-\frac{2}{7}\)
\(=\frac{110}{7}\)
a) \(\dfrac{5}{9}+\dfrac{13}{7}+\dfrac{15}{13}+\dfrac{8}{7}+\dfrac{4}{9}+\dfrac{11}{13}\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{13}{7}+\dfrac{8}{7}\right)+\left(\dfrac{15}{13}+\dfrac{11}{13}\right)\)
\(=1+3+2\)
\(=6\)
b) \(\dfrac{3}{5}+\dfrac{18}{9}+\dfrac{4}{10}\)
\(=0,6+2+0,4\)
\(=\left(0,6+0,4\right)+2\)
\(=1+2\)
\(=3\)
\(a\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{13}{7}+\dfrac{8}{7}+\dfrac{15}{13}+\dfrac{11}{13}=\dfrac{9}{9}+\dfrac{21}{7}+\dfrac{26}{13}=1+3+2=6\)
tính bằng cách thuận tiện nhất
a,4/15 x 7/8 + 4/15 x 2/9
b, 13/29 x 23/11 -13/19 x 8/11 - 13/19 x 4/11
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20
=(1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+20+10
=20+20+20+20+20+20+20+20+20+20+10
=20×10+10
=200+10
=210
= ( 1 + 19 ) + ( 2 + 18 ) + ( 3 + 17 ) + (4 + 16 ) + ( 5 + 15 ) + ( 6 + 14 ) + ( 7 + 13 ) + ( 8 + 12) + ( 9+ 11 ) + 20
= 20 + 20 + 20 +20 + 20 +20 + 20 + 20 + 20 +20
= 20 x 10
= 200
tui đầu tiên đó
(19+1)+(18+2)+(17+3)+(16+4)+(15+5)+(14+6)+(13+7)+(12+8)+(11+9)+20)+10
=20+20+20+20+20+20+20+20+20+20=(20*10)+10
=210
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20
=(1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+(10+20)
=20+20+20+20+20+20+20+20+20+30
=(20 x 9) + 30
=180 + 30
=210
tk nha
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
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a) 7/15+14/15+13/15=(7/15+13/15)+14/15=20/15+14/15=34/15
b)13/4+4/9+7/8=(13/4+7/8)+4/9=(26/8+7/8)+4/9=33/8+4/9=297/72+32/72=329/72
a. 7/15 + 14/15 + 13/15
= 7+14+13/ 15
= 34/15
b.13/4 + 4/9 + 7/8
= [ 13/4 + 7/8 ] + 4/9
= [ 26/8 + 7/8 ] + 4/9
= 33/8 + 4/9
= 297/72 + 32/72
= 329/72