1. \(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)
A nguyên nên \(3⋮n-2\). Vậy \(n-2\in\left(1,-1,3,-3\right)\Rightarrow n\in\left(3,1,5,-1\right)\)thì A nguyên.
2. a,Ta cần CM \(\frac{a}{b}< \frac{a+c}{b+c}\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow ab+ac< ab+bc\Rightarrow ac< bc\)(luôn đúng)
Suy ra điều phải chứng minh.
b, Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
Có:(suy ra từ phần a) \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Vậy \(1< \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\)
BẤM ĐÚNG CHO MÌNH, KO THÌ LẦN SAU KO GIÚP NỮA
Để \(A=\frac{n+1}{n-2}\)có giá trị nguyên => n + 1 chia hết cho n-2
\(=>\left(n-2\right)+3⋮\)\(n-2\)
Mà \(\left(n-2\right)⋮\)\(n-2\)
\(=>3⋮\)\(n-2\)
\(=>n-2\inƯ\left(3\right)=\){1;-1;3;-3}
Ta có bảng :
n-2 | 1 | -1 | 3 | -3 |
n | 3 | 1 | 5 | -1 |
Vậy \(n\in\){3;1;5;-1} để \(A=\frac{n+1}{n-2}\in Z\)
Câu 2 :
a)Vì \(a< b=>\frac{a}{b}< 1\)
Ta so sánh : \(\frac{a}{b}\)và \(\frac{a+c}{b+c}\)
Ta có : \(\frac{a}{b}=\frac{a.\left(b+c\right)}{b.\left(b+c\right)}=\frac{ab+ac}{b.\left(b+c\right)}\)
\(\frac{a+c}{b+c}=\frac{b.\left(a+c\right)}{b.\left(b+c\right)}=\frac{ab+bc}{b.\left(b+c\right)}\)
Vì a,b,c \(\in\)N* mà a<b => \(ab+ac< ab+bc=>\frac{ab+ac}{b.\left(b+c\right)}< \frac{ab+bc}{b.\left(b+c\right)}=>\frac{a}{b}< \frac{a+c}{b+c}\left(đpcm\right)\)
b) Ta có : \(\frac{a}{a+b}>\frac{a}{a+b+c}\)
\(\frac{b}{b+c}>\frac{b}{a+b+c}\)
\(\frac{c}{a+c}>\frac{c}{a+b+c}\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1\) \(\left(1\right)\)
Vì a,b,c \(\in\)N* mà a<b => \(\frac{a}{a+b}< 1\)=>\(\frac{a}{a+b}< \frac{a+c}{a+b+c}\)
Tương tự như vậy ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}+\frac{c+b}{a+b+c}=\frac{a+c+b+a+c+b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)=>1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}< 2\left(đpcm\right)\)