Đặt \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+...+\dfrac{1}{99.1}}\)

Đặt \(B=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\) ; \(C=\dfrac{1}{1.99}+\dfrac{1}{3.97}+...+\dfrac{1}{99.1}\)

=> \(A=\dfrac{B}{C}\) (*)

Ta có : \(C=\dfrac{1}{1.99}+\dfrac{1}{3.97}+...+\dfrac{1}{99.1}\)

\(\Rightarrow100C=\dfrac{100}{1.99}+\dfrac{100}{3.97}+...+\dfrac{100}{99.1}\)

\(\Rightarrow100C=\dfrac{99+1}{1.99}+\dfrac{97+3}{3.97}+...+\dfrac{99+1}{99.1}\)

\(\Rightarrow100C=1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+...+\dfrac{1}{99}+1\)

\(\Rightarrow100C=2.\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)\)

\(\Rightarrow50C=1+\dfrac{1}{3}+...+\dfrac{1}{99}\)=B , kết hợp (*)

\(\Rightarrow A=\dfrac{B}{C}=\dfrac{50C}{C}=50\)

TH1: \(\dfrac{1}{4}-x=0\Rightarrow x=\dfrac{1}{4}\)

TH2: \(x+\dfrac{2}{5}=0\Rightarrow x=-\dfrac{2}{5}\)

Vậy \(x\in\left\{\dfrac{1}{4};-\dfrac{2}{5}\right\}\)

c) Trường hợp 1:

 \(\dfrac{1}{4}-x=0\)

\(x=\dfrac{1}{4}-0\)

\(x=\dfrac{1}{4}\)

TH2:

\(x+\dfrac{2}{5}=0\)

\(x=0\)

Vậy x = 0; 0

\(F\left(x\right)=ax^3+4x\left(x^2-x\right)-4x+8=\left(a+4\right)x^3-4x^2-4x+8\)

\(G\left(x\right)=x^3-4x\left(bx+1\right)+c-3=x^3-4bx^2-4x+c-3\)

Để \(F\left(x\right)=G\left(x\right)\Rightarrow\left\{{}\begin{matrix}a+4=1\\4=4b\\c-3=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-3\\b=1\\c=11\end{matrix}\right.\)

   [(-0,25) - 3/4 ] : (-5) - 3 . (-1/2)2 + 1/5

= [(-1/4) - 3/4 ] : (-5) - 3 . 1/4 + 1/5

= (-1) : (-5) - 3 . 1/4 + 1/5

= 1/5 - 3/4 + 1/5

= 4/20 - 15/20 + 4/20

= -7/20

\(\left[\left(-0,25\right)-\dfrac{3}{4}\right]:\left(-5\right)-3.\left(-\dfrac{1}{2}\right)^2+\dfrac{1}{5}\\ =\left[-\dfrac{1}{4}-\dfrac{3}{4}\right]:\left(-5\right)-3.\dfrac{1}{4}+\dfrac{1}{5}\\ =\left(-1\right):\left(-5\right)-\dfrac{3}{4}+\dfrac{1}{5}\\ =\dfrac{1}{5}-\dfrac{3}{4}+\dfrac{1}{5}\\ =-\dfrac{7}{20}\)

   7/3 + [(-5/6) + (-2/3)]

= 7/3 + [(-5/6) + (-4/6)]

= 14/6 + (-9/6)

= 5/6

7/3 + [(-5/6) + (-2/3)]

= 7/3 + [(-5/6) + (-4/6)]

= 7/3 + (-9/6)

= 14/6 + (-9/6)

= 5/6

Ta có : \(Y=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{2006^2}\right)\left(1-\dfrac{1}{2007^2}\right)\)

\(\Rightarrow Y=\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{2006^2-1}{2006^2}.\dfrac{2007^2-1}{2007^2}\)

\(\Rightarrow Y=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{2006.2008}{2007.2007}=\dfrac{\left(1.2.....2006\right).\left(3.4.....2008\right)}{\left(2.3.....2007\right)\left(2.3.....2007\right)}=\dfrac{1.2018}{2007.2}=\dfrac{2018}{4014}\)

\(Y=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2007^2}\right)\)

\(=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.....\dfrac{2007^2-1}{2007^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.....\dfrac{2006.2008}{2007^2}\)

\(=\dfrac{\left(1.3\right).\left(2.4\right).....\left(2006.2008\right)}{2^2.3^2.....2007^2}\)

\(=\dfrac{\left(1.2.3.....2006\right).\left(3.4.5.....2008\right)}{\left(2.3.4.....2007\right).\left(2.3.4.....2007\right)}\)

\(=\dfrac{1.2008}{2007.2}=\dfrac{1004}{2007}\)

1/4 = 0,25 

Chiều rộng là:

   0,25 x 2/5 = 0,1 ( km)

Chu vi là:

     ( 0,25 + 0,1 ) x 2 = 0,7 ( km)

Diện tích là:

    0,25 x 0,1 = 0,025 ( km2 )

                  Đáp số:.................

 1/4 = 0,25   Chiều rộng là:

   0,25 x 2/5 = 0,1 ( km)

Chu vi là:

     ( 0,25 + 0,1 ) x 2 = 0,7 ( km)

Diện tích là:

    0,25 x 0,1 = 0,025 ( km2 )

                  Đáp số: Cv: 0,7  km

                                  S : 0,025 km2