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a)\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (1)
\(A=\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (có 2011 số hạng)
nếu ta trừ một vào từng số hạng được tử số giống nhau
\(A-2011=\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{4024}{2012}-1\right)\)
\(A-2011=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}=2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(A-2011+2012=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)công 2012 hai vế
\(A+1=VP=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\left(1\right)\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\left(2\right)\)
Chia cả hai vế (2) cho: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\Rightarrow503x=2012\)
\(x=\frac{2012}{503}\)
mình cố tình đặt A phân ra cho bạn dẽ hiểu: Nếu ko từ vế phải =1+2011+2012(1/2+...1/2012) =2012(1+1/2+...+1/2012) luôn không dài vậy
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
1/
\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)
\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
Phương trình đã cho tương đương:
\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=4\)
2/
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)
\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)
3/
Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=2.\frac{n+1}{n+2}
bạn bấm vào đúng 0 sẽ ra kết quả
mình làm bài này rồi
Ta có: Tử là:
B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\) (2013 số hạng 1)
=\(\left(\frac{1}{2013}+1\right)+\left(\frac{2}{2012}+1\right)+...+\left(\frac{2012}{2}+1\right)+\left(1\right)\)
=\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)
=\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014
bấm vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm bạn ạ
Phương trình đã cho tương đương với :
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2012=2012\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Tìm x theo như toán lớp 6 nha
\(x-2013=0\)
\(\Leftrightarrow\)\(x=2013\)
ta có pt
<=>\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1=0\)
<=>\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\Leftrightarrow x-2013=0\Leftrightarrow x=2013\)
^_^
Ta có phương trình :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+....+\frac{x-2012}{1}=2012\)
Ta thấy phương trình đã cho tương ứng với phương trình :
\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)+2012=2012\)
\(\Rightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Rightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+....+1\right)=0\)
Mặt khác \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)
Do đó \(\Rightarrow x-2013=0\Rightarrow x=2013\)
Do vậy \(x=2013\)thoả mãn phương trình ban đầu
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2000}+.....+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+........+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+......+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+......+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+.....+1\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+....+1\ne0\)
Vậy ...
\(\Leftrightarrow x=2013\)
\(\Leftrightarrow x-2013=0\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)\cdot503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}\)
\(\Leftrightarrow503x=\frac{1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)
\(\Leftrightarrow503x=\frac{\frac{2014}{2}-1+\frac{2015}{3}-1+...+\frac{4024}{2012}-1+2012}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)
\(\Leftrightarrow503x=\frac{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}+2012}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)
\(\Leftrightarrow503x=\frac{2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=\frac{2012}{503}\)