tính nhanh:
_____1_____+ ___1_____+____1______+......+_____1____
2*3 3*4 4*5 10*11
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\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
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) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
a) 5 – 4 – 3 – 2 – 1 + 11 + 12 + 13 + 14 + 15
= (5 + 15) + (14 – 4) + (13 – 3) + (12 – 2) + (11 – 1)
= 20 + 10 + 10 + 10 + 10
= 60
b) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +11
= (1 + 3 + 5 + 7 + 9 + 11) – 2 – 4 – 8 – 10
= 36 – 24
= 12
a) 5 – 4 – 3 – 2 – 1 + 11 + 12 + 13 + 14 + 15
= (5 + 15) + (14 – 4) + (13 – 3) + (12 – 2) + (11 – 1)
= 20 + 10 + 10 + 10 + 10
= 60
b) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 +11
= (1 + 3 + 5 + 7 + 9 + 11) – 2 – 4 – 8 – 10
= 36 – 24
= 12
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
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
Ta có \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
A=\(\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10x11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)