26/81 + 4/27
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= 26/81 + \(\dfrac{4X3}{27X3}\)
= 26/81 + 12/ 81
= \(\dfrac{26+12}{81}=\dfrac{38}{81}\)
B={x\(\in\)N|x=3k; 1<=k<=4}
C={x\(\in\)N|x=4*a2; 1<=a<=5}
D={x\(\in\)N|x=9*a2;1<=a<=4}
E={x\(\in\)N|x=4k; 0<=x<=4}
G={x\(\in\)N|x=(-3)^k; 1<=k<=4}
a: 2/3+3/4=11/12
9/4+3/5=45/20+12/20=57/20
2/5+4/7=14/35+20/35=34/34
3/5+4/3=9/15+20/15=29/15
b: 3/12+1/4=2/4=1/2
4/25+3/5=4/25+15/25=19/25
26/81+4/27=26/81+12/81=38/81
5/64+7/8=5/64+56/64=61/64
a: Ta có: \(81^{125}=3^{500}\)
\(27^{130}=3^{390}\)
mà 500>390
nên \(81^{125}>27^{130}\)
a)
\(15+27+9+85+73\)
\(=\left(15+85\right)+\left(27+73\right)+39\)
\(=200+39\)
\(=329\)
b)
\(54.36+54.27+54.37\)
\(=54\left(36+27+37\right)\)
\(=54.100\)
\(=5400\)
c)
\(168-3^7:3^4+2013^0\)
\(=168-3^3+1\)
\(=168-27+1\)
\(=142\)
d)
\(3^4.15+81.26+27.59.3\)
\(=81.15+81.26+81.59\)
\(=81.\left(15+26+59\right)\)
\(=81.100\)
\(=8100\)
Tính giá trị của các biểu thức sau :
a.15 + 27 + 39 + 85 + 73
= ( 15 + 85 ) + ( 27 + 73 ) + 39
= 100 + 100 + 39
= 200 + 39
= 239
b. 54 .36 + 54 . 27 + 54 . 37
= 54 . ( 36 + 27 + 37 )
= 54 . 100
= 5400
c.168 - 37 : 34 + 20130
= 168 - 33 + 1
= 168 - 27 + 1
= 141 + 1
= 142
đ. 34 . 15 + 81 . 26 + 27 . 59 . 3
= 34 . 15 + 34 . 26 + 34 . 59
= 34 . ( 15 + 26 + 59 )
= 81 . 100
= 8100
Lời giải:
$3^2\equiv -1\pmod 5\Rightarrow 3^{25}=(3^2)^{12}.3\equiv (-1)^{12}.3\equiv 3\pmod 5$
$\Rightarrow 2-3^{25}\equiv 2-3\equiv -1\pmod 5$
$\Rightarrow 2-3^{25}\not\vdots 5$.
Mà $3^{27}$ cũng không chia hết cho 5.
$\Rightarrow A$ không chia hết cho 5. Do đó $A$ không thể chia hết cho 15.
(\(\dfrac{2}{3}\) + \(\dfrac{8}{9}\) + \(\dfrac{26}{27}\) + \(\dfrac{80}{81}\) + \(\dfrac{242}{243}\)) : y = 5
Đăt A = \(\dfrac{2}{3}\) + \(\dfrac{8}{9}\) + \(\dfrac{26}{27}\) + \(\dfrac{80}{81}\) + \(\dfrac{242}{243}\)
3A = 2 + \(\dfrac{8}{3}\) + \(\dfrac{26}{9}\) + \(\dfrac{80}{27}\) + \(\dfrac{242}{81}\)
3A - A = 2 + \(\dfrac{8}{3}\) + \(\dfrac{26}{9}\) + \(\dfrac{80}{27}\) + \(\dfrac{242}{81}\) - \(\dfrac{2}{3}\)-\(\dfrac{8}{9}\)-\(\dfrac{26}{27}\)-\(\dfrac{80}{81}\)-\(\dfrac{242}{243}\)
A x (3 - 1) = 2 - \(\dfrac{242}{243}\)+ (\(\dfrac{8}{3}\) - \(\dfrac{2}{3}\))+(\(\dfrac{26}{9}\) - \(\dfrac{8}{9}\))+(\(\dfrac{80}{27}\)-\(\dfrac{26}{27}\))+(\(\dfrac{242}{81}\)-\(\dfrac{80}{81}\))-\(\dfrac{242}{243}\)
A x 2 = 2 - \(\dfrac{242}{243}\) + 2 + 2 + 2 + 2
A x 2 = (2 + 2 + 2 +2 + 2) - \(\dfrac{242}{243}\)
A x 2 = 2x5 - \(\dfrac{242}{243}\)
A x 2 = 10 - \(\dfrac{242}{243}\)
A x 2 = \(\dfrac{2188}{243}\)
A = \(\dfrac{2188}{243}\) : 2
A = \(\dfrac{1094}{243}\)
\(\dfrac{1094}{243}\) : y = 5
y = \(\dfrac{1094}{243}\) : 5
y = \(\dfrac{1094}{1215}\)
=38/81 nha emm =)
bấm mày tính đi máa