47568 : 4 ; 2 x 135
2005 - ( 175 ; 5 - 35 ) x 92
( 427 x 54 + 427 + 427 x 45 ) : 5
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Ta có \(3756a7b⋮2\)
\(\Rightarrow b\in\left\{0;2;4;6;8\right\}\)
4......4.......4......4=68
4......4........4......4=48
4.......4........4......4=24
4.......4........4.......4=20
4........4........4.......4=5
(x+2)(x+5)(x+3)(x+4)-24=(X^2+7x+10)(x^2+7x+12)-24. Đặt x^2+7x+10=t đa thức đã cho trở thành t*(t+2)-24=t^2+2t-24=t^2+6t-4t-24=t*(t+6)-4(t+6). =(t+6)(t-4).
Ta có:
\(\frac{\left(2^4+4\right).\left(6^4+4\right).\left(10^4+4\right).\left(14^4+4\right)}{\left(4^4+4\right).\left(8^4+4\right).\left(12^4+4\right).\left(16^4+4\right)}\)
\(=\frac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right).\left(9^2+1\right).\left(11^2+1\right).\left(13^2+1\right).\left(15^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right).\left(9^2+1\right).\left(11^2+1\right).\left(13^2+1\right).\left(15^2+1\right).\left(17^2+1\right)}\)
\(=\frac{1^2+1}{17^2+1}=\frac{1}{145}\)
\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)\left(14^4+4\right)}{\left(4^4+4\right)\left(8^4+4^4\right)\left(12^4+4\right)\left(16^4+4\right)}\)
\(=\frac{4\left(2^4+6^4+10^4+14^4\right)}{4\left(4^4+8^4+12^4+16^4\right)}\)
\(=\frac{4.76848}{4.90624}\)
\(=\frac{307392}{362496}=\frac{1601}{1888}\)
Làm:
47568 : 4 : 2 x 135
= 11892 : 270
= 44 , 0444444444
2005 - ( 175 : 5 - 35) x 92
= 2005 - ( 35 - 35 ) x92
=2005 - 0 x92 = 2005 - 0
= 2005
( 427 x 54 + 427 + 427 x 45 ) : 5
= ( 427 x 54 + 427 x 1 + 427 x45 ) : 5
=( 427 x ( 54 +1 + 45 ) ) : 5
= (427 X 100 ) : 5
= 42700 : 5
= 8540
( K Gìum , nếu sai thì hok cần k , ko thik thì cx chẳng cần k )