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\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~
\(\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{\left(n-1\right)^2-1}{\left(n-1\right)^2}\right)\left(\frac{n^2-1}{n^2}\right)\)
=\(\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}.\frac{\left(4-1\right)\left(4+1\right)}{4^2}...\frac{\left(n-2\right)n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
=\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{\left(n-2\right).n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}=\frac{1}{2}.\frac{n+1}{n}=\frac{1}{2}+\frac{1}{2n}>\frac{1}{2}\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)-2
=\(\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+1\right]:\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=\(\left(\frac{101}{2}+\frac{101}{3}+\frac{101}{4}+....+\frac{101}{100}+\frac{101}{101}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=\(101\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=99
Giả sử cả 3 bđt trên đều đúng, như vậy \(a\left(1-a\right).b\left(1-b\right).c\left(1-c\right)>\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\)
Mặt khác vì \(0< a,b,c< 1\) nên:
\(0< a\left(1-a\right)=-a^2+a-\frac{1}{4}+\frac{1}{4}=\frac{1}{4}-\left(a-\frac{1}{2}\right)^2\le\frac{1}{4}\)
Tương tự \(0< b\left(1-b\right)\le\frac{1}{4}\) và \(0< c\left(1-c\right)\le\frac{1}{4}\)
Suy ra \(a\left(1-a\right).b\left(1-b\right).c\left(1-c\right)\le\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\) (vô lý)
Vậy phải có ít nhất 1 bđt sai
đặt \(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}+\frac{1}{102}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}+\frac{1}{102}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{51}\right)\)
\(A=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
Khó quá taa=))
Hmm