Question

1tìm \(n\in Z\)để \(A=\frac{n+1}{n-2}\left(n\ne2\right)\)có giá trị nguyên

2 cho \(a,b,c\in N\)*  và a<b

Hãy chứng tỏ \(\frac{a}{b}< \frac{a+c}{b+c}\)và \(1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)

Answer 1

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

Answer 2

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 \(A=\frac{n+1}{n-2}\in Z\)

Answer 3

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 \(\left(1\right);\left(2\right)=>1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}< 2\left(đpcm\right)\)