Giả sử  $ a \le b \le c \le d $ . Chứng minh rằng  phương trình:   $ p(x - a)(x - c) + q(x - b)(x - d) = 0 $ luôn luôn có nghiệm với mọi $p$ và $q$.
Đặt $ f\left( x \right) = p(x - a)(x - c) + q(x - b)(x - d) $
Nếu  $ p = q = 0 $:
Ta có:  $ f\left( x \right) = 0 $
Nếu $ p = 0;q \ne 0 $ : $ f\left( x \right) = q(x - b)(x - d) $
Phương trình  $ f\left( x \right) = 0 $ có 2 nghiệm: $ {x_1} = b;{x_2} = d $
Nếu $ q = 0;p \ne 0 $ : $ f\left( x \right) = p(x - a)(x - c) $
Phương trình  $ f\left( x \right) = 0 $  có 2 nghiệm : $ {x_3} = a;{x_4} = c $
Nếu  $ p \ne 0;q \ne 0 $
Ta có : $ f\left( b \right).f\left( d \right) = {p^2}(b - a)(b - c)(d - a)(d - c) < 0 $  $  \Rightarrow  $ phương trình có hai nghiệm  $ {x_1};{x_2} $ mà 1 nghiệm thuộc khoảng (b,d)
Đó là đpcm.

Thẻ

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