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1/10+2/9+3/8+4/7+5/6+1/6+3/7+5/8+7/9+9/10
=(1/10+9/10)+(2/9+7/9)+(3/8+5/8)+(4/7+3/7)+(5/6+1/6)
=1+1+1+1+1
=5
Bài 7:
Số phần kẹo Hùng đã cho Hà và Hồng là:
\(\dfrac{2}{7}+\dfrac{1}{7}=\dfrac{3}{7}\left(phần\right)\)
Hùng còn lại số phần của gói kẹo là:
\(\dfrac{6}{7}-\dfrac{3}{7}=\dfrac{3}{7}\left(phần\right)\)
1:
2 3/4
5 6/5
3 3/9
7 6/8
2:
1/3 + 2/3 + (3/4 + 1/4) = 2
=2
= 4 5/10
Ta có:
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
= 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= (36 – 36) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
= 0
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= (27 – 27) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7
=0
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7 Nếu đúng thì k cho mình nhé bạn!
\(\frac{9}{10}\)+\(\frac{7}{9}\)+\(\frac{5}{8}\)+\(\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)\)\(+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=1+1+1+1+1\)
\(=5\)
9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
= ( 9/10 + 1/10 ) + ( 7/9 + 2/9 ) + ( 5/8 + 3/8 ) + ( 3/7 + 4/7 ) + ( 3/5 + 2/5 )
= 1 + 1 + 1 + 1 + 1
= 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)
= \(1+1+1+1+1\)
= \(1\times5\)
= \(5\)
Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :
A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)
A = 1 + 1 + 1 + 1 + 1
A = 5
( 1/10 + 2/9 + 3/8 + 4/7 + 5/6 + 1/6 + 3/7 + 5/8 + 7/9 + 9/10 ) * 48
= [ ( 1/10 + 9/10 ) + ( 2/9 + 7/9 ) + ( 3/8 + 5/8 ) + ( 4/7 + 3/7 ) + ( 5/6 + 1/6 ) ] * 48
=( 1 + 1 + 1 + 1 + ) *48
=5 * 48
= 240
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)\(=\frac{9}{10}+\frac{1}{10}+\frac{7}{9}+\frac{2}{9}+\frac{5}{8}+\frac{3}{8}+\frac{3}{7}+\frac{4}{7}+\frac{3}{5}+\frac{2}{5}\)
\(=\left[\frac{9}{10}+\frac{1}{10}\right]+\left[\frac{7}{9}+\frac{2}{9}\right]+\left[\frac{5}{8}+\frac{3}{8}\right]+\left[\frac{3}{7}+\frac{4}{7}\right]+\left[\frac{3}{5}+\frac{2}{5}\right]\)
\(=1+1+1+1+1\)
\(=5\)
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= 1 + 1 + 1 + 1 + 1
= 5
a, 9/1 * 3/8 + 3/8 * 9/7
b, 4/5 * 7/10 - 1/9 * 4/5
C1:nhân chia trước cộng trừ sau
C2:a,3,8x(9,1+9,7)=3,8x18,8=?
b,b, 4/5 * 7/10 - 1/9 * 4/5=4/5x(7/10-1/9)=?
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