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Ta có: \(\frac{n}{2n+1}=\frac{3n}{6n+3}\)
Vì 3n < 3n + 1 nên \(\frac{3n}{6n+3}<\frac{3n+1}{6n+3}\)
Vậy \(\frac{n}{2n+1}<\frac{3n+1}{6n+3}\)
Ta có:
n/2n + 1 = 3n/6n + 3
3n/6n + 3 < 3n + 1/6n + 3
=>n/2n + 1 <3n + 1/6n + 3
Thanks!
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\(\frac{n}{2n+1}\)=\(\frac{3.n}{3.\left(2n+1\right)}\)=\(\frac{3n}{6n+3}\)
Vì 6n+3=6n+3;3n<3n+1 nên \(\frac{n}{2n+1}\)<\(\frac{3n+1}{6n+3}\)
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\(A=\dfrac{n}{2n+1}=\dfrac{n\left(6n+3\right)}{\left(2n+1\right)\left(6n+3\right)}\dfrac{6n^2+3n}{\left(2n+1\right)\left(6n+3\right)}\)
\(B=\dfrac{3n+1}{6n+3}=\dfrac{\left(3n+1\right)\left(2n+1\right)}{\left(6n+3\right)\left(2n+1\right)}=\dfrac{6n^2+5n+1}{\left(6n+3\right)\left(2n+1\right)}\)
Lại có :
\(6n^2+3n< 6n^2+5n+1\)
\(\Leftrightarrow A< B\)
A=n2n+1=n(6n+3)(2n+1)(6n+3)6n2+3n(2n+1)(6n+3)
B=3n+16n+3=(3n+1)(2n+1)(6n+3)(2n+1)=6n2+5n+1(6n+3)(2n+1)
Lại có :
6n2+3n<6n2+5n+1
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a) Ta thấy :
27 chia hết cho 3
6n = 3.2.n chia hết cho 2.n
Vậy n = 0; 1; 2; 3; 4; 5; 6; ... hay n = mọi số tự nhiên .
b) 2n + 5 chia hết cho 3n + 1
2n + 4 + 1 chia hết cho 2n + n + 1
Vì 2n + 1 chia hết cho 2n + 1 nên 4 chia hết cho n
Ư(4) = 1; 2; 4
Vậy n = 1; 2; 4
Cấm COPY
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3n/3x 2n+3 =n/2n+1
n/2n+1=3n/3x2n+3=3n/6n+3<3n/6n+2
n/2n+1<3n+1/6n+2
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có Q=3n+1/6n+3
=3n/(6n+3)+1/(6n+3)
=3n/3.(2n+1)+1/6n+3
=n/3n+1+1/6n+3