Question

[Phuc An Tran]

câu b bài 84 ạ giúp với ;^;giúp

Answer 1

[Nguyễn Thanh Hằng]

a, Ta có :

A = \(\dfrac{a}{a+b}\) + \(\dfrac{b}{b+c}\) + \(\dfrac{c}{c+a}\) (a; b; c thuộc N*)

Ta có :

\(\dfrac{a}{a+b}\) < 1 => \(\dfrac{a}{a+b}\) < \(\dfrac{a+c}{a+b+c}\)

Tương tự :

\(\dfrac{b}{b+c}\) < \(\dfrac{b+a}{b+c+a}\)

\(\dfrac{c}{c+a}\) < \(\dfrac{c+b}{c+a+b}\)

=> A < \(\dfrac{2\left(a+b+c\right)}{a+b+c}\)= 2 (1)

Mặt khác :

\(\dfrac{a}{a+b+c}\) < \(\dfrac{a}{a+b}\)

\(\dfrac{b}{a+b+c}\) < \(\dfrac{b}{b+c}\)

\(\dfrac{c}{a+b+c}\) < \(\dfrac{c}{c+a}\)

=> \(\dfrac{a+b+c}{a+b+c}\) < A

1 < A (2)

Từ (1) và (2) => 1 < A < 2

=> A ko thể là 1 số nguyên ( do 1 và 2 là 2 số nguyên liên tiếp)

Câu b tương tự nha bn!!

Chúc bn học tốt!!

Answer 2

[Nguyễn Huy Tú]

b) Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\\\dfrac{b}{b+c+d}< \dfrac{b+a}{a+b+c+d}\\\dfrac{c}{c+d+a}< \dfrac{c+b}{a+b+c+d}\\\dfrac{d}{d+a+b}< \dfrac{b+c}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \dfrac{a+d}{a+b+c+d}+\dfrac{b+a}{a+b+c+d}+\dfrac{c+b}{a+b+c+d}+\dfrac{b+c}{a+b+c+d}\)

\(=\dfrac{2a+2b+2c+2d}{a+b+c+d}=\dfrac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

\(\Rightarrow B< 2\) (1)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\\\dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}\\\dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}\\\dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>\dfrac{a+b+c+d}{a+b+c+d}=1\)

\(\Rightarrow B>1\) (2)

Từ (1) và (2) \(\Rightarrow1< B< 2\)

\(\Rightarrow B\notin Z\left(đpcm\right)\)

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