Giải các phương trình sau:
1/ \(\frac{x+1}{2011}\)+\(\frac{x+2}{2010}\)=\(\frac{x+3}{2009}+\frac{x+4}{2008}\)
2/ (3x-2)2-x(9x-2)=24
Giải phương trình
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
tự làm nốt~
kudo shinichi làm sai ở chỗ:
\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé
Giải phương trình sau:
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
giải phương trình: \(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2009}{2}+\frac{x+2010}{1}\)\(=\left(-2010\right)\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
cách bạn giải bài này là gì vậy ?
Giải phương trình
\(\frac{1}{2}\left(\frac{2x-2}{2009}+\frac{2x}{2010}+\frac{2x+2}{2011}\right)=\frac{33}{10}-\left(\frac{x+1}{2011}+\frac{x-1}{2009}+\frac{x}{2010}\right)\)
\(a.\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(b.\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức
\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)
\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}-\frac{x+2010}{2013}-\frac{x+2010}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow\)\(x=2010\)
Vậy \(x=2010\)
Chúc bạn học tốt ~
a.\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)\left(\frac{x-4}{2014}+1\right)\)
=> \(\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
=>\(\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
=>x+2010=0 (vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\)\(\ne0\)
=> x = -2010
Vậy x= -2010
Câu b làm tương tự nhưng -1 nhé !
giải phương trình
\(\frac{3-x}{2009}-\frac{2-x}{2010}+\frac{1-x}{2011}=-1.\)
Bài này đề sai thì phải ??
\(\frac{3-x}{2009}-\frac{2-x}{2010}+\frac{1-x}{2011}=-1\)
\(\Leftrightarrow\frac{3-x}{2009}-\frac{2-x}{2010}+\frac{1-x}{2011}+1=0\)
\(\Leftrightarrow\left(\frac{3-x}{2009}+1\right)-\left(\frac{2-x}{2010}+1\right)+\left(\frac{1-x}{2011}+1\right)=0\)
\(\Leftrightarrow\frac{2012-x}{2009}-\frac{2012-x}{2010}+\frac{2012-x}{2011}=0\)
\(\Leftrightarrow\left(2012-x\right)\left(\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2011}\right)=0\)
Mà \(\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2011}\ne0\)
\(\Leftrightarrow2012-x=0\)
\(\Leftrightarrow x=2012\)
Vậy \(x=2012\)
giải các phương trình
a) \(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
b) \(\frac{x+1}{2009}+\frac{x+2}{2008}=\frac{x+2007}{3}+\frac{x+2006}{4}\)
a) \(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
\(\frac{8-6x-4+x}{10}=\frac{5x+10}{10}\)
\(4-5x=5x+10\)
\(4-5x-5x-10=0\)
\(-6-10x=0\)
\(\Rightarrow x=\frac{-3}{5}\)
Vậy....
\(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)
\(\Leftrightarrow\)\(\frac{2.\left(4-3x\right)}{10}-\frac{4-x}{10}=\frac{5.\left(x+2\right)}{10}\)
\(\Rightarrow\) 2.( 4 - 3x ) - 4 + x = 5.( x + 2 )
\(\Leftrightarrow\)8 - 6x - 4+ x = 5x + `10
\(\Leftrightarrow\)-6x + x - 5x = -8 + 4 + 10
\(\Leftrightarrow\) -10x = 6
\(\Leftrightarrow\)\(x=\frac{-3}{5}\)
Vậy phương trình có nghiệm là: \(x=\frac{-3}{5}\)
b ) \(\frac{x+1}{2009}+\frac{x+2}{2008}=\frac{x+2007}{3}+\frac{x+2006}{4}\)
\(\Leftrightarrow\) \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1\)\(=\frac{x+2007}{3}+1+\frac{x+2006}{4}+1\)
\(\Leftrightarrow\)\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}\)\(=\frac{x+2007}{3}+\frac{3}{3}+\frac{x+2006}{4}+\frac{4}{4}\)
\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{3}+\frac{x+2006}{4}\)
\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{3}-\frac{x+2010}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+2010=0\) ( Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\ne0\))
\(\Leftrightarrow\) \(x=-2010\)
Vậy phương trình có nghiệm là: x = -2010
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)tìm x
\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
Giải phương trình: \(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
Ta có :
\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)
\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)
\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)
Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)
\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài
Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)