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Ta có x < y ; m > 0
=> \(\frac{a}{m}< \frac{b}{m}\)
=> a < b (vì m > 0)
Lại có x = \(\frac{a}{m}=\frac{2a}{2m}=\frac{a+a}{2m}< \frac{a+b}{2m}=y\)(vì a < b nên a + a < a + b)
=> x < z (1)
Mặt khác \(y=\frac{b}{m}=\frac{2b}{2m}=\frac{b+b}{2m}>\frac{a+b}{2m}=z\)(vì b > a nên b +b > b + a)
=> y > z (2)
Từ (1) và (2) => x < z < y (đpcm)
\(\frac{a}{b}>1\Rightarrow a>b>m\)
Ta có:
\(\frac{a-m}{b-m}=\frac{ab-bm}{\left(b-m\right).b}\)
\(\frac{a}{b}=\frac{ab-am}{\left(b-m\right).b}\)
\(am>bm\left(a>b\right)\)
\(\Rightarrow ab-bm>ab-am\)
\(\Rightarrow\frac{a-m}{b-m}>\frac{a}{b}\left(1\right)\)
\(\frac{a+m}{b+m}=\frac{ab+bm}{\left(b+m\right).b}\)
\(\frac{a}{b}=\frac{ab+am}{\left(b+m\right).b}\)
\(bm< am\left(b< a\right)\)
\(\Rightarrow ab+bm< ab+am\)
\(\Rightarrow\frac{a+m}{b+m}< \frac{a}{b}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{a-m}{b-m}>\frac{a}{b}>\frac{a+m}{b+m}\)
+ Do a/b > 1
=> a > b
=> a.m > b.m
=> a.b - a.m < a.b - b.m
=> a.(b - m) < b.(a - m)
=> a/b < a-m/b-m (1)
Do a/b > 1
=> a > b
=> a.m > b.m
=> a.m + a.b > b.m + a.b
=> a.(b + m) > b.(a + m)
=> a/b > a+m/b+m (2)
Từ (1) và (2) => a-m/b-m > a/b > a+m/b+m
Ủng hộ mk nha ☆_☆^_-
ta có
a,\(\frac{a}{b}< 1\Leftrightarrow a< b\Leftrightarrow a+m< b+m\)
vì \(a+m< b+m\)
nên \(\frac{a+m}{b+m}< 1\)
b,Ta có \(a+b>1\Leftrightarrow a+m>b+m\)
Vì \(a+m>b+m\)
nên \(\frac{a+m}{b+m}>1\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Biến đổi vế 2 :
\(\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}\)( quy đồng )
\(=\frac{bc+ac+ab}{abc}\)
Ta có :
\(=\frac{\left(a+b+c\right)\left(bc+ac+ab\right)}{abc}\)
\(=\frac{abc+abc+abc}{abc}\)\(=3\)
→ ( a + b + c ) = 3
Ta có : 3 . 3 = 9 => ĐPCM
a, M=1/1.2+1/2.3+...+1/49.50
M=1−1/2+1/2−1/3+...+1/49−1/50
M=1−1/50<1
Vậy M<1
\(a,\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}< 1\)
\(=>M< 1\)
\(\frac{a}{b}\)< 1 <=> a < b <=> a.m < b.m <=> ab + a.m < ab + b.m
<=> a(b + m) < b(a + m)
<=> \(\frac{a}{b}\)< \(\frac{a+m}{b+m}\)
Bài 1:
Có: \(\frac{a}{a+b}>\frac{a}{a+b+c};\frac{b}{b+c}>\frac{b}{b+c+a};\frac{c}{a+c}>\frac{c}{a+c+b}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\\ \Rightarrow A>\frac{a+b+c}{a+b+c}\Rightarrow A>1\left(1\right)\)
Lại có: \(\frac{a}{a+b}< 1\Rightarrow\frac{a}{a+b}< \frac{a+c}{a+b+c};\frac{b}{b+c}< 1\Rightarrow\frac{b}{b+c}< \frac{b+a}{b+c+a};\frac{c}{a+c}< 1\Rightarrow\frac{c}{a+c}< \frac{c+b}{a+c+b}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}< \frac{a+c}{a+b+c}+\frac{b+a}{b+c+a}+\frac{c+b}{a+c+b}\\ \Rightarrow A< \frac{a+c+b+a+c+b}{a+b+c}\Rightarrow A< \frac{2a+2b+2c}{a+b+c}\Rightarrow A< \frac{2\left(a+b+c\right)}{a+b+c}\Rightarrow A< 2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow1< A< 2\left(đpcm\right)\)
Bài 2 ;
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{91.94}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{91}-\frac{1}{94}\)
= \(1-\frac{1}{94}< 1\)
Vậy ........(đpcm )