tính bằng cách hợp lý
\(3^{2018}\). [ \(3^{20}\) : (\(3^8\) . \(3^{11}\)) +2 .3]
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A=(1+2+3+...+200) : (6+8+10+...+34)
A=\(\frac{\left[\left(200-1+1\right).\left(200+1\right)\right]}{2}:\frac{\left[\left(34-6\right):2+1\right].\left(34+6\right)}{2}\)
A=20100 : 300
A=67.
B=( 41x66+34x66):(3+7+11+...+79)
B=[(41+34).66]:\(\frac{\left[\left(79-3\right):4+1\right].\left(79+3\right)}{2}\)
B=(75.66):820
B=4950:820
B=\(\frac{495}{82}\)
13-12+11+10-9+8-7-6+5-4+3+2-1
=13-(12+1) -(11+2)-(10+3)-(9+4)-(8+5) +(7+6)
=13-13-13-13-13-13+13
=0-0-0-0-0-0+13
=0+13
=13
sửa lại đề bài đi : Tính hợp lí:13 - 12 + 11 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 +2 - 1
:A
\(\frac{11}{8}+0,25-\left(\frac{1}{3}\right)^2+20\%-1\)
\(=\frac{11}{8}+\frac{2}{8}-\frac{1}{9}+\frac{1}{5}-1\)
\(=\frac{13}{8}-\left(\frac{1}{9}-\frac{1}{5}+1\right)\)
\(=\frac{13}{8}-\left(\frac{5}{45}-\frac{9}{45}+\frac{45}{45}\right)\)
\(=\frac{13}{8}-\frac{41}{45}\)
\(=\frac{585}{360}-\frac{328}{360}\)
\(=\frac{257}{360}\)
a)\(\left(316-25-19^2\right):56+19\cdot5\)
\(=\left(316-25-361\right):56+95\)
\(=-70\cdot56+95=-3825\)
b) \(\left(3^5\cdot3^8:3^{11}-2^3\right)+2018=\left(3^{13}:3^{11}-8\right)+2018=\left(3^2-8\right)+2018=1+2018=2019\)
\(a,\left[316-25-19^2\right]:56+19.5\)
\(=\left[291-361\right]:56+95\)
\(=-70:56+95\)
\(=-1,25+95\)
\(=93.75\)
\(b,\left[3^5.3^8:3^{11}-2^3\right]+2018\)
\(=\left[3^{13}:3^{11}-8\right]+2018\)
\(=\left[3^2-8\right]+2018\)
\(=\left[9-8\right]+2018\)
\(1+2018=2019\)
Hc tốt nha ^^
`a, 4/9 + 2/5 + 5/9= (4/9 +5/9) + 2/5= 1 + 2/5= 5/5 + 2/5= 7/5`
`b, 6/13 + 5/8 + 7/13=(6/13 + 7/13) + 5/8=1+5/8= 8/8 + 5/8= 13/8`
`c, 5/11 + 6/25 + 66/121 = ( 5/11 + 66/121 ) + 6/25=1 + 6/25 = 25/25 + 6/25 = 31/25`
`d, 1/2 + 1/3 + 1/4 + 1/6 + 1/12 + 1/24= (1/2 + 1/4) +(1/3 + 1/6) + (1/12 + 1/24)= 3/4 + 1/2 + 1/8= 6/8 + 4/8 + 1/8= 11/8`
`@ yl`
\(3^{2018}\left(3^{20}:\left(3^8\cdot3^{11}\right)+2\cdot3\right)=3^{2018}\left(3^{20}:3^{19}+2\cdot3\right)\)
\(=3^{2018}\left(3+2\cdot3\right)=3^{2018}\cdot9=3^{2018}\cdot3^2\)
\(=3^{2020}\)