cho b>0. So sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+1}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: a < b => a + 1 < b + 1
b) Ta có: a < b => a - 2 < b - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
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Để so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+1}\), ta đi so sánh hai số \(a\left(b+1\right)\)và \(b\left(a+1\right)\).
Xét hiệu:
\(a\left(b+1\right)-b\left(a+1\right)=ab+a-\left(ab+b\right)=a-b\)
Ta có 3 trường hợp, với điều kiện b > 0:
Trường hợp 1: Nếu \(a-b=0\Leftrightarrow a=b\)thì:
\(a\left(b+1\right)-b\left(a+1\right)=0\Leftrightarrow a\left(b+1\right)=b\left(a+1\right)\)
\(\Leftrightarrow\frac{a\left(b+1\right)}{b\left(a+1\right)}=\frac{b\left(a+1\right)}{a\left(b+1\right)}\Leftrightarrow\frac{a}{b}=\frac{a+1}{b+1}\)
Trường hợp 2: Nếu \(a-b< 0\Leftrightarrow a< b\)thì:
\(a\left(b+1\right)-b\left(a+1\right)< 0\Leftrightarrow a\left(b+1\right)< b\left(a+1\right)\)
\(\Leftrightarrow\frac{a\left(b+1\right)}{b\left(a+1\right)}< \frac{b\left(a+1\right)}{a\left(b+1\right)}\Leftrightarrow\frac{a}{b}< \frac{a+1}{b+1}\)
Trường hợp 3: Nếu \(a-b>0\Leftrightarrow a>b\)thì:
\(a\left(b+1\right)-b\left(a+1\right)>0\Leftrightarrow a\left(b+1\right)>b\left(a+1\right)\)
\(\Leftrightarrow\frac{a\left(b+1\right)}{b\left(a+1\right)}>\frac{b\left(a+1\right)}{a\left(b+1\right)}\Leftrightarrow\frac{a}{b}>\frac{a+1}{b+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A
Vậy A > B
b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A
Vậy A < B.
NHỚ K CHO MK VỚI NHÉ !!!!!!!!
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Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1
Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)
10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1
Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)
Từ (1) và (2) => 10A < 10B
=> A < B
Tk mk nha
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)
\(\Rightarrow\)\(A,B< 1\)
Ta có:
\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)
\(\Rightarrow A>B\)
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
sai đề rồi bạn.\(\frac{a}{b}>\frac{a+c}{b+c}\) với \(a>b\) mới đúng nha.
Ta có:\(A=\frac{10^{17}+1}{10^{16}+1}>\frac{10^{17}+1+9}{10^{16}+1+9}=\frac{10^{17}+10}{10^{16}+10}=\frac{10\left(10^{16}+1\right)}{10\left(10^{15}+1\right)}=\frac{10^{16}+1}{10^{15}+1}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: a-1/a = a/a - 1/a = 1 - 1/a < 1
b+1/b = b/b + 1/b = 1 + 1/b > 1
=> a-1/a < 1 < b+1/b
Vậỵ a-1/a < b+1/b
Xét 3 trường hợp :
+) Nếu b > a thì \(\frac{a}{b}=\frac{b-m}{b}=\frac{b}{b}-\frac{m}{b}=1-\frac{m}{b}\)
\(\frac{a+1}{b+1}=\frac{b-m+1}{b+1}=\frac{b+1-m}{b+1}=\frac{b+1}{b+1}-\frac{m}{b+1}=1-\frac{m}{b+1}\)
Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1-\frac{m}{b}< 1-\frac{m}{b+1}\)hay \(\frac{a}{b}< \frac{a+1}{b+1}\)
+) Nếu a = b thì \(\frac{a}{b}=1\)
\(\frac{a+1}{b+1}=1\)nên\(\frac{a}{b}=\frac{a+1}{b+1}\)
+) Nếu a > b thì \(\frac{a}{b}=\frac{b+m}{b}=\frac{b}{b}+\frac{m}{b}=1+\frac{m}{b}\)
\(\frac{a+1}{b+1}=\frac{b+m+1}{b+1}=\frac{b+1}{b+1}+\frac{m}{b+1}=1+\frac{m}{b+1}\)
Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1+\frac{m}{b}>1+\frac{m}{b+1}\)hay \(\frac{a}{b}>\frac{a+1}{b+1}\)
Ta có :
\(\frac{a}{b}=\frac{a\left(b+1\right)}{b\left(b+1\right)}=\frac{ab+a}{b^2+b}\)
\(\frac{a+1}{b+1}=\frac{b\left(a+1\right)}{b\left(b+1\right)}=\frac{ab+b}{b^2+b}\)
Từ 2 ý trên , ta xét từng trường hợp sau :
a < b thì \(\frac{a}{b}< \frac{a+1}{b+1}\)
a > b thì \(\frac{a}{b}>\frac{a+1}{b+1}\)
a = b thì \(\frac{a}{b}=\frac{a+1}{b+1}\)