Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)=0+1+1+1+1\)
\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{2011}=4\)
\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=4\)
\(\Leftrightarrow x+2015=4:\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x+2015=2012,499379\)
\(\Leftrightarrow x=2012,499379-2015\)
\(\Leftrightarrow x=-2,5006118\)
P/s : Số xấu quá !
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010+18}{6}=0\)
\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010}{6}+3=0\)
\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2010}{6}\right)=0\)
\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+6\right)=0\)
Vì :\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
=> x + 2010 = 0
=> x = -2010
b) \(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\frac{x-3}{2011}+\frac{x-2}{2012}-\frac{x-2012}{2}-\frac{x-2011}{3}=0\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-2012}{2}-1\right)-\left(\frac{x-2011}{3}-1\right)=0\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)
\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)
=> x - 2014 = 0
=> x = 2014
c) \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Rightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
\(\Rightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Rightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì :\(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
=> x + 66 = 0
=> x = -66
a)
PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)
<=> x - 2013 = 0
<=> x = 2013
KL: ...
b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)
<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)
KL: ...
a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)
nên x-2013=0
hay x=2013
Vậy: Tập nghiệm S={2013}
b) Ta có: \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\)(2)
Từ (1) và (2) suy ra (x+6)(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)
Vậy: Tập nghiệm S={-6;5}
\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)
\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy x = 2014
có 2 kho thóc.kho thứ nhất lớn hơn kho thứ 2 100tấn.Nếu chuyển từ kho thú nhất sang kho 2 60tấn thì kho thóc thứ nhất bằng 12/13 số thóc thứ hai .Tính số thóc mỗi thùng
