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\(a,\left(-2\right)^{300}=2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(\left(-3\right)^{200}=3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(\Leftrightarrow\left(-3\right)^{200}>\left(-2\right)^{300}\)
\(b,A=2^{19}.27^3+15.4^9.9^4\)
ta có
\(\frac{1}{15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\frac{1}{35}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(\frac{1}{63}=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}\right)\)
......................................
\(\frac{1}{143}=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}\right)\)
Cộng hết lại: \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)
B = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(\Rightarrow B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(\Rightarrow B=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+\frac{1.2}{9.11.2}+\frac{1.2}{11.13.2}\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{10}{39}\)
\(\Rightarrow B=\frac{5}{39}\)
Vậy \(B=\frac{5}{39}\)
Gọi dãy là A ta có :
A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13
A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )
A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )
A = 1/2 . ( 1/3 - 1/13 )
A = 1/2 . 10/39
A = 5/39
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
=\(\frac{1}{2}.\frac{10}{39}\)
=\(\frac{5}{39}\)
1/15+1/35+1/63+1/99+1/143
=1/2x(1/15+1/35+1/63+1/99+1/143)
=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)
=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
Đặt: A=1/15+1/35+1/63+1/99+1/43
Chắc chắn A<1.
ta co 1/15<1/5 ; 1/35 <1/5 ;1/63<1/5;1/99<1/5 ;1/143<1/5 suy ra 1/15+1/35+1/63+1/99+1/143<1/5+1/5+1/5+1/5+1/5
tuong duong 1/15+1/35+1/63+1/99+1/143<1