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\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4-3-1}{12}\right)\)
\(=\left(\dfrac{66}{111}+\dfrac{2}{33}+\dfrac{15}{117}\right).0\)
\(=0\)
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\\ =\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4-3-1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot0\\ =0\)
a) 22 + 3 . 23 = 22 + 3 . 8 = 22 + 24 = 46
b) 41 . 67 + 33 . 41 = 41 . ( 67 + 33 ) = 41 . 100 = 4100
c) 322 . 5 - [ 622 : 4 + 2 . ( 11 - 8) ] = 322 . 5 - [ 622 : 4 + 2 . 3 ] = 322 . 5 - [ 155,5 + 6 ] = 322 . 5 - 161,5 = 1610 + 161,5 = 1448,5
\(A=\frac{595959}{454545}-\frac{14141414}{15151515}\)
\(A=\frac{595959:10101}{454545:10101}-\frac{14141414:1010101}{15151515:1010101}\)
\(A=\frac{59}{45}-\frac{14}{15}\)
\(A=\frac{59}{45}-\frac{42}{45}=\frac{17}{45}\)
\(B=\frac{65+891+135+909}{731-47+69-253}\)
\(B=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
\(B=\frac{200+1800}{800-300}=\frac{2000}{500}=4\)
1)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Sửa đề chút nha
\(\frac{x}{2}=\frac{1}{1.2.3}+....+\frac{1}{98.99.100}\)
Ta có công thức tổng quát \(\frac{1}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{2}\left(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right)\)
\(\Rightarrow\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Áp dụng vào tổng ta có
\(\frac{x}{2}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow x=\frac{4949}{4950}\)
a) 135 + 360 + 65 + 40 = (135 + 65) + (360 + 40) = 200 + 400 = 600.
b) 463 + 318 + 137 + 22 = (463 + 137) + (318 + 22) = 600 + 340 =940.
c) Nhận thấy 20 + 30 = 50 = 21 + 29 = 22+ 28 = 23 + 27 = 24 + 26.
Do đó 20 + 21 + 22 + ...+ 29 + 30
= (20+ 30) + (21 + 29) + (22 + 28) + (23 + 27) + (24 + 26) + 25
= 5 . 50 + 25 = 275.
BÀI LÀM:
a, 135 + 360 + 65 + 40 = ( 135 + 65 ) + ( 360 + 40 ) = 200 + 400 = 600
b, 463 + 318 + 137 + 22 = ( 463 + 137 ) + ( 318 + 22 ) = 600 + 340 = 940
c, 20 + 21 + 22 + ... + 29 + 30
= ( 20 + 30 ) + ( 21 + 29 ) + ( 22 + 28 ) + ( 23 + 27 ) + ( 24 + 26 ) + 25
= 50 . 5 + 25 = 275
1) \(5\frac{2}{7}\).\(\frac{8}{11}\)+\(5\frac{2}{7}.\frac{5}{11}+5\frac{2}{7}.\frac{-2}{11}\)
= \(5\frac{2}{7}\).[\(\frac{8}{11}+\frac{5}{11}+\frac{-2}{11}\)]
= \(5\frac{2}{7}\).1
=\(5\frac{2}{7}\)
2)
5 2/7.8/9+5 2/7.5/11-5 2/7.2/11
=5 2/7.(8/11+5/11+2/11)
=5 2/7.15/11
=240/77.
(67/111+2/33-15/117).(1/3-1/4-1/12)
=811/1221.1/12
=811/14652
Ta có:
\(A=\frac{67^{2016}}{67^{2016}-11}=1+\frac{11}{67^{2016}-11}\)
\(B=\frac{67^{2016}+13}{67^{2016}+2}=1+\frac{11}{67^{2016}+2}\)
Vì \(67^{2016}-11< 67^{2016}+2\) nên \(\frac{11}{67^{2016}-11}>\frac{11}{67^{2016}+2}\Rightarrow1+\frac{11}{67^{2016}-11}>1+\frac{11}{67^{2016}+2}\)
Vậy A > B
\(C=\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)x\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(\Rightarrow C=\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)x\left(\frac{1}{12}-\frac{1}{12}\right)\)
\(\Rightarrow C=\left(\frac{67}{11}+\frac{2}{33}-\frac{15}{117}\right)x0\)
\(\Rightarrow C=0\)
67 +135 + 33 = (67 + 33) +135 = 100 +135 - 235.