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\(\frac{x-7}{2005}+\frac{x-6}{2006}=\frac{x-5}{2007}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-7}{2005}-1+\frac{x-6}{2006}-1=\frac{x-5}{2007}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\). Do \(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\ne0\)
\(\Rightarrow x=2012\)
\(\frac{x-7}{2005}+\frac{x-6}{2006}=\frac{x-5}{2007}+\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-7}{2005}-1\right)+\left(\frac{x-6}{2006}-1\right)=\left(\frac{x-5}{2007}-1\right)+\left(\frac{x-4}{2008}-1\right)\)
\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)
Mà \(\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)\ne0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
Vậy \(x=2012\)
1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=2010\)
\(\Rightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Rightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\ne0\)
nên \(x-2010=0\Leftrightarrow x=2010\)
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
1 Ta có x -24 = y
Suy ra x - y = 24
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
x/7 = y/3 = x-y/7-3 =24/4=6
suy ra x= 42
y = 18
1) \(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{13}+\frac{x+4}{12}\)
\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{13}-\frac{x+16}{12}=0\)
\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{13}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x=-16\)
2)3)4) tương tự
Gợi ý : 2) cộng 3 vào cả hai vế
3)4) cộng 2 vào cả hai vế
5) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=-3\)
\(\Leftrightarrow\frac{x+21}{20}+\frac{x+21}{19}+\frac{x+21}{18}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}\right)=0\)
\(\Leftrightarrow x=-21\)
6) sửa VT = 4 rồi tương tự câu 5)
\(\left(\frac{x-7}{2005}-1\right)+\left(\frac{x-6}{2006}-1\right)=\left(\frac{x-5}{2007}-1\right)+\left(\frac{x-4}{2008}-1\right)\)
\(\Leftrightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)
\(\left(x-2012\right).\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)
\(\text{vì }\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)\ne0\Rightarrow x-2012=0\Rightarrow x-2012\)