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b ) Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{101.103}\)
\(\Rightarrow A=\frac{5}{2}\left(\frac{5}{1}-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+....+\frac{5}{101}-\frac{5}{103}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(5-\frac{5}{103}\right)\)
A=\(\frac{2}{1}\cdot3+\frac{2}{3}\cdot5+\frac{2}{5}\cdot7+\frac{2}{7}\cdot9\)
A=\(\frac{3}{1}\cdot2+\frac{5}{3}\cdot2+\frac{7}{5}\cdot2+\frac{9}{7}\cdot2\)
A=\(2\cdot\left(3+\frac{5}{3}+\frac{7}{5}+\frac{9}{7}\right)\)
A=\(2\cdot\frac{772}{105}\)
A=\(\frac{1544}{105}\)
A=\(14\frac{74}{105}\)
450x0,45 + 1,5x30x3 + 5x9x2,5
= 450x0,45 + 4,5x30 + 45x2,5
= 450x0,45 + 0,45x300 + 0,45x250
= 0,45x(450 + 300 + 250) = 0,45x1000 = 450
AI lấy đêf kiểm tra toán giữa kì 1 không
ai nhanh thì mình sẽ cho
\(\frac{2}{9}\cdot\frac{5}{7}+\frac{2}{7}\cdot\frac{2}{9}\)
\(=\frac{2}{9}\cdot\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=\frac{2}{9}\)
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
\(\left(\dfrac{1}{2}+\dfrac{1}{4}\right)\times\left(9\times5-5\times9\right)=\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\times0=0\)
( 3/5 + 5/9) + 2/5
= ( 3/5 + 2/5 ) + 5/9
= 1 + 5/9
= 14/9
(3/5+5/9)+2/5
=(2/5+3/5)+5/9
=1+5/9
=14/9