Tính nhanh:
7/10+3:10+2/5+1:10
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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
\(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}\)
\(=\frac{1+2+3+4+5+6+7+8+9}{10}\)
\(=\frac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}=\frac{10.4+5}{10}=\frac{45}{10}=\frac{9}{2}\)
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10
= { 1 + 9 } + { 2 + 8 } + { 3 + 7 } + { 4 + 6 } + 5
= 10 + 10 + 10 + 10 + 5
= 45
\(=\dfrac{1+2+3+4+5+6+7+8+9}{10}\\ =\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\\ =\dfrac{10+10+10+10+5}{10}=\dfrac{45}{10}=\dfrac{9}{2}\)
\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.
2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.
3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
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
1+2+3+4+5+6+7+8+9+10+11+12+13-1-2-3-4-5-6-7-8-9-10-11-12
=(1-1)+(2-2)+(3-3)+(4-4)+(5-5)+(6-6)+(7-7)+(8-8)+(9-9)+(10-10)+(11-11)+(12-12)+13
= 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 13
= 13
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 - 1 - 2 - 3 - 4 - 5 - 6 - 7- 8 - 9 - 10
= (1+12) + (2+11) + (3+10) + ( 4+9) + (5+8) + (6+7) + 13 - (1+11) - ( 2+10) - ( 3+9) - (4+8) - (5+7) + 6 + 12
= 13 * 7 - 12 * 6 + 6
= 13
Ai k mk mk k lại
Đáp án đề thi học sinh giỏi lớp 2 môn Toán
Bài 1: Tính nhanh
a. 10 – 9 + 8 – 7 + 6 – 5 + 4 – 3 + 2 – 1
= (10 – 9) + (8 – 7) + (6 – 5) + (4 – 3) + (2 – 1)
= 1 + 1 + 1 + 1 + 1
= 5
b. 1 + 3 + 5 + 7 + 9 + 10 + 8 + 6 + 4 + 2 + 0
= (0 +10) +(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5
= 10 + 10 + 10 + 10 + 10 + 5
= 55
a ) 10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1
= ( 10 - 9 ) + ( 8 - 7 ) + ( 6 - 5 ) + ( 4 - 3 ) + ( 2 - 1 )
= 1 + 1 + 1 + 1 + 1
= 5
b ) 1 + 3 + 5 + 7 + 9 + 10 + 8 + 6 + 4 + 2 + 0
= ( 10 + 0 ) + ( 1 + 9 ) + ( 2 + 8 ) + ( 3 + 7 ) + ( 4 + 6 ) + 5
= 10 + 10 + 10 + 10 + 10 + 5
= 55
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=\frac{15}{15}+\frac{7}{7}\)
\(=1+1\)
\(=2\)
b) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
\(\dfrac{7}{10}+3:10+\dfrac{2}{5}+1:10\\ =\left(\dfrac{7}{10}+\dfrac{2}{5}\right)+\left(3:10+1:10\right)\\ =\dfrac{11}{10}+\left(3+1\right)\times\dfrac{1}{10}\\ =\dfrac{11}{10}+4\times\dfrac{1}{10}\\ =\dfrac{11}{10}+\dfrac{2}{5}\\ =\dfrac{3}{2}\)
\(\dfrac{7}{10}+3:10+\dfrac{2}{5}+1:10\\ =\dfrac{7}{10}+\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{1}{10}\\ =\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\left(\dfrac{2}{5}+\dfrac{1}{10}\right)\\ =\dfrac{10}{10}+\left(\dfrac{4}{10}+\dfrac{5}{10}\right)=\dfrac{10}{10}+\dfrac{5}{10}=\dfrac{15}{10}=\dfrac{3}{2}\)