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a,3/5*7/9+11/9*3/5-3/10=3/5*(7/9+11/9)-3/10
=3/5*18/9-3/10
=3/5*2-3/10
=6/5-3/10
=12/10-3/10
=9/10
b,14/13:1/2+3/13*2-2*7/13=14/13*2+3/13*2-2*7/13
=2*(14/13+3/13-7/13)
=2*10/13
=20/13
a/ ( 7/9 + 11/9 ) x 3/5 - 3/10
= 2 x 3/5 - 3/10
= 6/5 - 3/10
= 9/10
b/ 14/13 : 1/2 + 3/13 x 2 - 2 x 7/13
14/13 x 2/1 + 3/13 x 2 - 2 x 7/13
2 x ( 14/13 + 3/13 - 7/13 )
2 x 10/13
20/13
A)\(\frac{3}{5}\times\frac{7}{9}+\frac{11}{9}\times\frac{3}{5}-\frac{3}{10}\\ =\frac{3}{5}\times\left(\frac{7}{9}+\frac{11}{9}\right)-\frac{3}{10}\\ =\frac{3}{5}\times\frac{18}{9}-\frac{3}{10}=\frac{3\times2}{5}-\frac{3}{10}\)
\(=\frac{6}{5}-\frac{3}{10}=\frac{12}{10}-\frac{3}{10}=\frac{9}{10}\)
\(\dfrac{2}{3}+\dfrac{7}{13}-\dfrac{17}{9}+\dfrac{19}{13}-\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)-\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=\dfrac{9}{3}+\dfrac{26}{13}-\dfrac{18}{9}\)
\(=3+2-2\)
\(=3\)
(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
\(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}=\left(\frac{3}{5}+\frac{2}{5}\right)\left(\frac{6}{11}+\frac{16}{11}\right)\left(\frac{7}{13}+\frac{19}{13}\right)=1+2+2=5\)
Ta có:3/5+2/5=5/5=1; 6/11+16/11=22/11=2; 7/13+19/13=26/13=2
kết quả là 1+2+2=5
a, =4/18+5/13+7/9+21/13
=4/18+5/13+14/18+21/13
=[4/18+14/18]+ [5/13+21/13]
=1+1/2
=3/2
Bài 3
a,26/100+0,009+41/100+0,24
0,26+0,09+0,41+0,24
(0,26+0,24)+(0,09+0,41)
0,5+0,5
=1
b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4
(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)
30+1+1+1+1
=34
Bài 4,5 khó quá mik ko bít lamf^^))
Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1
\(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)
b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)
Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)