giúp
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!!
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) \(\Rightarrow1< B< 2\)
\(\Rightarrow B\notin Z\left(đpcm\right)\)
Vậy...
Bài 84:
a) \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)
Ta có:
\(\left\{\begin{matrix}\frac{a}{a+b}>\frac{a}{a+b+c}\left(1\right)\\\frac{b}{b+c}>\frac{b}{a+b+c}\left(2\right)\\\frac{c}{c+a}>\frac{c}{a+b+c}\left(3\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1\)\((*)\)
Lại có:
\(\left\{\begin{matrix}\frac{a}{a+b}< \frac{a+c}{a+b+c}\left(3\right)\\\frac{b}{b+c}< \frac{b+a}{a+b+c}\left(4\right)\\\frac{c}{c+a}< \frac{c+b}{a+b+c}\left(5\right)\end{matrix}\right.\)
Từ \(\left(3\right);\left(4\right)\) và \(\left(5\right)\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=2\)\((*)(*)\)
Từ \((*)\) và \((*)(*)\)
\(\Rightarrow1< A< 2\)
Vậy \(A\) không là số nguyên (Đpcm)
b) \(B=\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)
Ta có:
\(\Rightarrow\left\{\begin{matrix}\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\left(1\right)\\\frac{b}{b+c+d}>\frac{b}{a+b+c+d}\left(2\right)\\\frac{c}{c+d+a}>\frac{c}{a+b+c+d}\left(3\right)\\\frac{d}{d+a+b}>\frac{d}{a+b+c+d}\left(4\right)\end{matrix}\right.\)
\(\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1\)
\(\Rightarrow B>1\)\((*)\)
Lại có:
\(\Rightarrow\left\{\begin{matrix}\frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\left(1\right)\\\frac{b}{b+c+d}< \frac{b+c}{a+b+c+d}\left(2\right)\\\frac{c}{c+d+a}< \frac{c+a}{a+b+c+d}\left(3\right)\\\frac{d}{d+a+b}< \frac{d+b}{a+b+c+d}\left(4\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right);\left(3\right)\) và \(\left(4\right)\)
\(\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+d}{a+b+c+d}+\frac{b+c}{a+b+c+d}+\frac{c+a}{a+b+c+d}+\frac{d+b}{a+b+c+d}=2\)
\(\Rightarrow B< 2\)\((*)(*)\)
Từ \((*)\) và \((*)(*)\)
\(\Rightarrow1< B< 2\)
Vậy \(B\) không là số nguyên (Đpcm)
Giúp mk nha mấy bn, bik bài nào thì giúp, ko bik thì thoy! Giúp mk mấy bài khoanh
