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\(1-A=\frac{n^6-n^5}{n^6+1}=\frac{n^5\left(n-1\right)}{n^6+1}\)
\(1-B=\frac{n^5-n^4}{n^5+1}=\frac{n^4\left(n-1\right)}{n^5+1}=\frac{n^5\left(n-1\right)}{n^6+n}\)
Vì n6 +1 < n6 + n
=> 1 -A > 1-B
Hay A < B
\(1-A=1-\frac{n^5+1}{n^6+1}=\frac{n^5\left(n-1\right)}{n^6+1}\)
\(1-B=1-\frac{n^4+1}{n^5+1}=\frac{n^4\left(n-1\right)}{n^5+1}=\frac{n^5\left(n-1\right)}{n^6+n}\)
Vì n6 + 1 < n6 +n
=> 1 -A > 1-B
=> A < B
thầy nói đề sai rồi mà
phải là cm ƯCLN của a và b ko lớn hơn \(\sqrt{m+n}\)
Gọi \(gcd\left(m;n\right)=d\Rightarrow m=ad;n=bd\left(a,b\inℕ^∗\right)\) và \(\left(m;n\right)=1\)
Ta có:
\(\frac{m+1}{n}+\frac{n+1}{m}=\frac{m^2+m+n^2+n}{mn}=\frac{\left(a^2+b^2\right)d+\left(a+b\right)}{abd}\)
\(\Rightarrow a+b⋮d\Rightarrow a+b\ge d\Rightarrow d\le\sqrt{d\left(a+b\right)}=\sqrt{m+n}\)
Vậy ta có đpcm
Câu 2: n= 12
Do A=\(\frac{\left(2x2\right)^6x\left(2x3\right)^6}{3^6x2^6}=2^{12}\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};......;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-1\right)n}\)
Ta lại có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+.....+\frac{1}{n\left(n-1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{n^2}< 1\) (đpcm)
a) Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
\(\Rightarrow\)A < 1
b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(B=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{n^2}\right)\)
vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)
\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)
Vì n5 + 1 < n6 + 1
\(M=\frac{n^5+1}{n^6+1}< \frac{n^5+1+\left(n-1\right)}{n^6+1+\left(n-1\right)}=\frac{n^5+n}{n^6+n}=\frac{n\left(n^4+1\right)}{n\left(n^5+1\right)}=\frac{n^4+1}{n^5+1}=N\)
=> M < N
Ta có: \(N=\frac{n^4+1}{n^4+1}=1\) ( n > 1 )
\(M=\frac{n^5+1}{n^6+1}< 1\) ( do n > 1 )
\(\Rightarrow M< 1\) hay M < N
Vậy M < N