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a) Ta có:\(\dfrac{31}{67}>\dfrac{31}{73}\) (1)
\(\dfrac{31}{73}>\dfrac{29}{73}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{31}{67}>\dfrac{31}{73}>\dfrac{29}{73}\)
\(\Rightarrow\dfrac{31}{67}>\dfrac{29}{73}\)
Vậy:...............
\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)
Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B
\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)
Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C
\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)
=>
\(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+\dfrac{x+3}{97}+1+\dfrac{x+4}{96}+1=0\)
\(\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\) vậy \(x=-100\)
2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)
B=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+..+15^{96}+15^{100}\right)+\left(15^2+15^6+...+15^{98}+15^{102}\right)}\)
=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)+15^2.\left(1+15^{14}+15^8+...+15^{96}+15^{100}\right)}\)
\(\dfrac{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)\left(1+15^2\right)}\)
=\(\dfrac{1}{1+15^2}=\dfrac{1}{226}\)