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Nếu \(\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ab+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)
Vậy \(a< b\) thì \(\frac{a}{b}< \frac{a+m}{b+m}\) (đpcm)
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 \(A=\frac{n+1}{n-2}\in Z\)
Giải:
Theo đề bài ta có:
a:3 dư 2\(\rightarrow\)a+1 chia hết cho 3
a:5 dư 4\(\rightarrow\)a+1 chia hết cho 5
a: 7 dư 6\(\rightarrow\)a+1 chia hết cho 7
\(\Rightarrow a+1⋮3;5;7\rightarrowđpcm\)
\(a_{MIN};a+1⋮3;5;7\)
\(\Rightarrow a+1\in BCNN\left(3;5;7\right)=3.5.7=210\)
\(a=210-1=219\)
a:5 dư 3\(\Rightarrow2a-1⋮5\)
a:7 dư 4 \(\Rightarrow2a-1⋮7\)
a :11 dư 6 \(\Rightarrow2a-1⋮11\)
\(\Rightarrow2a-1⋮5;7;11\rightarrowđpcm\)
\(\Rightarrow2a-1\in BC\left(5;7;11\right)\)
\(BCNN\left(5;7;11\right)=5.7.11=385\)
\(B\left(385\right)=\left\{0;385;770;....\right\}\)
\(2a=\left\{1;386;771;....\right\}\)
\(a=\left\{\dfrac{1}{2};193;\dfrac{771}{2};....\right\}\)
\(100< a< 200\)
\(\rightarrow a=193\)
TA có
\(\frac{a}{b}-\frac{a+c}{b+c}=\frac{a\left(b+c\right)}{b\left(b+c\right)}-\frac{b\left(a+c\right)}{b\left(b+c\right)}\)
\(=\frac{ab+ac-ab-bc}{b\left(b+c\right)}=\frac{ac-bc}{b\left(b+c\right)}=\frac{c\left(a-b\right)}{b\left(b+c\right)}\)
vì a>b => a-b > 0 => c(a-b) > 0
=> \(\frac{c\left(a-b\right)}{b\left(b+c\right)}>0\)
\(=>\frac{a}{b}-\frac{a+c}{b+c}>0\)
\(=>\frac{a}{b}>\frac{a+c}{b+c}\)
=> đpcm
b) Ta có a+b < a+b+c ; b+c < a+b+c ; c+a < a+b+c
\(=>\frac{a}{a+b}>\frac{a}{a+b+c};\frac{b}{b+c}>\frac{b}{a+b+c};\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a+b+c}{a+b+c}=1\) (1)
Lại có
Áp dùng câu a ta có a< a+b ; b< b+c ; c<c+a
=> \(\frac{a}{a+b}< \frac{a+c}{a+b+c};\frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{c+a}< \frac{c+b}{a+b+c}\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{2\left(a+b+c\right)}{a+b+c}=2\) (2)
Từ (1) và (2) => dpcm
Vì \(a< b< c< d< m< n\)
\(\Rightarrow\hept{\begin{cases}a+c+m< 3a\\a+b+c+d+m+n< 6a\end{cases}}\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{3a}{6a}\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)
Bài giải
Ta có : \(a< b\text{ }\Rightarrow\text{ }2a< a+b\)
\(c< d\text{ }\Rightarrow\text{ }2c< c+d\)
\(m< n\text{ }\Rightarrow\text{ }2m< m+n\)
\(\Rightarrow\text{ }2a+2c+2m< \left(a+b+c+d+m+n\right)\) \(\Leftrightarrow\text{ }2\left(a+c+m\right)< \left(a+b+c+d+m+n\right)\)
\(\Rightarrow\text{ }\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
Do a < b < c < d < m < n
=> 2c < c + d
m< n => 2m < m+ n
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n)
Do đó :
(a + c + m)/(a + b + c + d + m + n) < 1/2(đcpcm)
Từ:\(\hept{\begin{cases}a< c\\c< d\\m< n\end{cases}}\Rightarrow a+c+m< c+d+n\)
\(\Rightarrow2\left(a+c+n\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
a,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 => ac < bc
\(\Rightarrow\frac{ab+ac}{b\left(b+c\right)}< \frac{ab+bc}{b\left(b+c\right)}\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\left(đpcm\right)\)
b, Ta có: \(\frac{a}{a+b}>\frac{a}{a+b+c}\)
\(\frac{b}{b+c}>\frac{b}{a+b+c}\)
\(\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\left(đpcm\right)\)
đã bảo em hoc dốt thì đừng ngu