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a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
a/ Áp dụng bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(\dfrac{10^{2011}+1}{10^{2012}+1}< 1\)
\(\Leftrightarrow\dfrac{10^{2011}+1}{10^{2012}+1}< \dfrac{10^{2011}+1+9}{10^{2012}+1+9}=\dfrac{10^{2011}+10}{10^{2012}+10}=\dfrac{10\left(10^{2010}+1\right)}{10\left(10^{2011}+1\right)}=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{10^{2011}+1}{10^{2012}+1}< \dfrac{10^{2010}+1}{10^{2011}+1}\)
Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
\(=>B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}\)
\(< \frac{10^{2012}+10}{10^{2013}+10}\)
\(< \frac{10.\left(10^{2011}+1\right)}{10.\left(10^{2012}+1\right)}\)
\(< \frac{10^{2011}+1}{10^{2012}+1}=A\)
=> B < A
Ủng hộ mk nha ^_-
Bài 1:
Ta có: \(A=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
Dễ thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow A=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< B=\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
Bài 2:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)
So sánh 2 phân số sau $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10
kick dzô chữ xanh là được!! OK
Ta có :
10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)
= \(\frac{10^{2012}+10}{10^{2012}+1}\)
= \(\frac{10^{2012}+1+9}{10^{2012}+1}\)
= \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)
= 1 - \(\frac{9}{10^{2012}+1}\)
10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)
= \(\frac{10^{2013}+10}{10^{2013}+1}\)
= \(\frac{10^{2013}+1+9}{10^{2013}+1}\)
= 1 - \(\frac{9}{10^{2013}+1}\)
Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\) nên 10.A > 10.B
=> A >B
Vậy ...........
Áp dụng tính chất \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(B=\dfrac{10^{2014}+1}{10^{2013}+1}< \dfrac{10^{2014}+1+9}{10^{2013}+1+9}\)
\(=\dfrac{10^{2014}+10}{10^{2013}+10}=\dfrac{10\left(10^{2013}+1\right)}{10\left(10^{2012}+1\right)}=\dfrac{10^{2013}+1}{10^{2012}+1}\)
\(\Rightarrow\dfrac{10^{2014}+1}{10^{2013}+1}< \dfrac{10^{2013}+1}{10^{2012}+1}\)
Hay \(A< B\)
Chết Chết! Nhầm rồi! Học lâu quá quên!
Sửa lại:
Cái tượng đề đầu tiên phải là:
Áp dụng tính chất \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\)
Còn phần sau thì chuyển dấu \(< \rightarrow>\)
Kết luận là \(A>B\)
Trước hết ta so sánh 10.A với 10.B từ đó ta \(\Rightarrow\) A < B. Ta có:
\(10.A=\dfrac{-10\left(10^{2011}+1\right)}{10^{2012}+1}=\dfrac{-\left(10^{2011}.10+10\right)}{10^{2012}+1}\)
\(=\dfrac{-\left(10^{2012}+10\right)}{10^{2012}+1}=\dfrac{-\left(10^{2012}+1\right)}{10^{2012}+1}-\dfrac{9}{10^{2012}+1}=-1-\dfrac{9}{10^{2012}+1}\)
Tương tự: \(10.B=-1-\dfrac{9}{10^{2013}+1}\)
Do \(10^{2013}+1>10^{2012}+1,\) nên \(\dfrac{-9}{10^{2013}+1}>\dfrac{-9}{10^{2012}+1}\)
Do đó \(10.A< 10.B,\) tức là \(A< B\)