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Lời giải:
a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)
Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$
Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$
Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$
b) Rõ ràng $10^{11}-1< 10^{12}-1$.
Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$
Áp dụng kết quả phần a:
$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$
\(10A=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-9}{10^{12}}-1\)
\(10B=\frac{10^{11}+1+9}{10^{11}+1}=\frac{10^{11}+9}{10^{11}}+1\)
ta có: \(A=\frac{10^{11}-1}{10^{12}-1}\)
\(\Rightarrow10.A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}\)\(=1-\frac{9}{10^{12}-1}< 1\)
ta có: \(B=\frac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow10.B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}\)\(=1+\frac{9}{10^{11}+1}>1\)
\(\Rightarrow10.A< 10.B\)
\(\Rightarrow A< B\)
Ta có :\(a=\dfrac{10^{11}-1}{10^{12}-1}\Rightarrow10a=\dfrac{10^{12}-10}{10^{12}-1}=\dfrac{10^{12}-1-9}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(b=\dfrac{10^{10}+1}{10^{11}+1}\Rightarrow10b=\dfrac{10^{11}+10}{10^{11}+1}=\dfrac{10^{11}+1+9}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Ta có : \(1-\dfrac{9}{10^{12}-1}\le1+\dfrac{9}{10^{11}+1}\) hay \(10a< 10b\Rightarrow a< b\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(A< \dfrac{10^{11}-1+11}{10^{12}-1+11}\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow A< B\)
ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}\\ 10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\\ =>10A< 1\\ B=\dfrac{10^{10}+1}{10^{11}+1}\\ 10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\\ =>10B>1\)
=> 10A<10B =>A<B
vậy A bé hơn B
Bạn ơi !
Hàng thứ 2 dưới lên phải viết là : Ta có : 10A < 10B => A < B
B/A= [(10^10 + 1)/(10^11 + 1)]/[(10^11 - 1)/(10^12 - 1)]
= [(10^12 - 1).(10^10 + 1)]/[(10^11 - 1).(10^11 + 1)]
= [(10^22 - 1) + (10^12 - 10^10) ]/((10^22 - 1)
= 1 + (10^12 - 10^10)/(10^22 - 1) > 1
=> B > A
Dấu "/" nghĩa là phân số nhé
Ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}\) \(B=\frac{10^{10}+1}{10^{11}+1}\)
\(10A=\frac{10^{12}-10}{10^{12}-1}\) \(10B=\frac{10^{11}+10}{10^{11}+1}\)
\(10A=\frac{10^{12}-1-9}{10^{12}-1}\) \(10B=\frac{10^{11}+1+9}{10^{11}+1}\)
\(10A=1-\frac{9}{10^{12}-1}\) \(10B=1+\frac{9}{10^{11}+1}\)
Ta thấy \(1-\frac{9}{10^{12}-1}< 1\) mà \(1+\frac{9}{10^{11}+1}>1\)
=> A < B
Vậy A < B
Ủng hộ mk nha !!! ^_^
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}\) theo công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{11}+10}{10^{12}+10}=\frac{10^{10}\left(10+1\right)}{10^{11}\left(10+1\right)}=\frac{10^{10}}{10^{11}}\)
\(\Rightarrow\frac{10^{10}}{10^{11}}=\frac{10^{10}\cdot10^{12}}{10^{11}\cdot10^{12}}=\frac{10^{22}}{10^{23}}\)
\(\Leftrightarrow A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}\)
Lại áp dụng công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}< \frac{10^{11}+1}{10^{12}+1}=B\)
\(\Leftrightarrow A< B\)
Hoặc \(A< \frac{10^{11}-1+2}{10^{12}-1+2}=\frac{10^{12}+1}{10^{12}+1}\)
..... (EZ)
ta có: \(A=\dfrac{10.10^{10}-1}{10.10^{11}-1}=\dfrac{10^{10}-1}{10^{11}-1}\)
so sánh: \(A=\dfrac{10^{10}-1}{10^{11}-1}\)và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow A< B\)
ta có :\(A=\frac{10^{11}-1}{10^{12}-1}=\frac{1}{10}=0,1\)
\(B=\frac{10^{10}+1}{10^{11}+1}=\frac{1}{10}=0,1\)
\(\Rightarrow A=\frac{1}{10}\)và \(B=\frac{1}{10}\)
Vậy \(A=B\)
Ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow A< B\)
Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}\)
\(\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\)
\(\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\)
\(\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Cách 2:
Ta có: \(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\Rightarrow1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)
\(\Rightarrow10A< 10B\Rightarrow A< B\)
Vậy A < B