21:(a+3)*4=100
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a) Ta có A = 1 + 4 + 42 + 43 + ... + 449
4A = 4 + 42 + 43 + 44 + ... + 450
4A - A = ( 4 + 42 + 43 + 44 + ... + 450 ) - ( 1 + 4 + 42 + 43 + ... + 449 )
3A = 450 - 1
A = \(\dfrac{4^{50}-1}{3}\)
Vì A = \(\dfrac{4^{50}-1}{3}\) < \(\dfrac{4^{100}}{3}\) = \(\dfrac{B}{3}\) nên A < \(\dfrac{B}{3}\)
b) Ta có A = 1 + 4 + 42 + 43 + ... + 449
= 1 + 4 + ( 42 + 43 + 44 ) + ( 45 + 46 + 47 ) + ... + ( 447 + 448 + 449 )
= 5 + 42( 1 + 4 + 42 ) + 45( 1 + 4 + 42 ) + ... + 447( 1 + 4 + 42 )
= 5 + 42 . 16 + 45 . 16 + ... + 447 . 16
= 5 + 21( 42 + 45 + ... + 447 )
Vì [ 21( 42 + 45 + ... + 447 )] ⋮ 21 nên A = 5 + 21( 42 + 45 + ... + 447 ) chia 21 dư 5
Vậy A chia 21 dư 5
đây là toán lớp 6 ư. Ròi xong tới công chuyện với me òi năm sau lên lớp 6
1a
75/100+18/21+19/32+1/4+3/21+13/32
= 3/4 +6/7+19/32+1/4+1/7+13/32
= (3/4+1/4)+(19/32+13/32)+(6/7+1/7)
= 1+1+1=3
1b
22/5+51/9+11/4+3/5+1/3+1/4
=22/5+17/3+11/4+3/5+1/3+1/4
=(22/5+3/5)+(17/3+1/3)+(11/4+1/4)
=25/5+18/3+12/4
=5+6+3
=14
a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105
b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50
a )
75/100 + 18/21 + 19/32 + 1/4 + 3/21 + 13/32
= 3/4 + 18/21 + 19/321 + 1/4 + 3/21 + 13/32
= ( 3/4 + 1/4 ) + ( 18/21 + 3/21 ) + ( 19/32 + 13/32 )
= 1 + 1 + 1
= 3
b )
4 và 2/5 + 5 và 6/9 + 2 và 3/4 + 1/4 + 1/3 + 3/5
= 22/5 + 51/9 + 11/4 + 1/4 + 1/3 + 3/5
= ( 22/5 + 3/5 ) + ( 51/9 + 1/3 ) + ( 11/4 + 1/4 )
= 25/5 + 54/9 + 12/4
= 5 + 6 + 3
= 14
a)\(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}=\frac{3}{4}+\frac{18}{21}+\frac{1}{4}+\frac{19}{32}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1=3\)
1a) 4^21=(4^2)^10.4=(....6)^10.4=(......6).4=(.......4)
b) 3^100=(3^4)^25=(.....1)^25=(.....1)
\(a)5\frac{3}{5}+1\frac{3}{4}+4\frac{1}{4}+3\frac{2}{5}\)
\(=\)\((5\frac{3}{5}+3\frac{2}{5})+(1\frac{3}{4}+4\frac{1}{4})\)
\(=[8+(\frac{3}{5}+\frac{2}{5})]+[5+(\frac{3}{4}+\frac{1}{4})]\)
\(=(8+1)+(5+1)\)
\(=9+6=15\)
\(b)\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=(\frac{3}{4}+\frac{1}{4})+(\frac{18}{21}+\frac{3}{21})+(\frac{19}{32}+\frac{13}{32})\)
\(=1+1+1=3\)
_Học tốt_