1,Tìm số hữu tỉ x biết\(\frac{x+4}{2005}+\frac{x+3}{2006}=\frac{x+2}{2007}+\frac{x+1}{2008}\)
2,tìm x biết:\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
Những câu hỏi liên quan
Tìm x biết
a)\(x^4-30x^2+31x-30=0\)
b)\(\left(x^2+x\right)^2+4\times\left(x^2+x\right)=12\)
c)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
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1) Cho a, b là 2 số hữu tỉ thỏa mãna^5+b^52a^2b^2CMR: 1 - ab là bình phương của 1 số hữu tỉ2) Cho x, y thỏa mãn left|x-2005right|+left|x-2006right|+left|y-2007right|+left|x-2008right|3 Tìm x, y.3) Cho Aleft(1-frac{1}{1+2}right).left(1-frac{1}{1+2+3}right).left(1-frac{1}{1+2+3+4}right)...left(1-frac{1}{1+2+3+...+n}right)với n-1 thừa số và Bfrac{n+2}{n}. Tìm frac{A}{B}
Đọc tiếp
1) Cho a, b là 2 số hữu tỉ thỏa mãn\(a^5+b^5=2a^2b^2\)
CMR: 1 - ab là bình phương của 1 số hữu tỉ
2) Cho x, y thỏa mãn \(\left|x-2005\right|+\left|x-2006\right|+\left|y-2007\right|+\left|x-2008\right|=3\) Tìm x, y.
3) Cho \(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+n}\right)\)
với n-1 thừa số và \(B=\frac{n+2}{n}\). Tìm \(\frac{A}{B}\)
Tìm x
a/\(\frac{x+7}{2003}+\frac{x+4}{2006}=\frac{x-1}{2011}+\frac{x-5}{2015}\)
b/\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c/\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
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<
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Giải phương trình:
1. \(\left(x^2+x\right)^2+4\left(x^2+x\right)^2=12\)
2. \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
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TÌM x BIẾT:a,frac{3}{left(x+2right)left(x+5right)}+frac{5}{left(x+5right)left(x+10right)}+frac{7}{left(x+10right)left(x+17right)}frac{x}{left(x+2right)left(x+17right)}với xnotin{-2;-5;-10;-17}b,frac{2}{left(x-1right)left(x-3right)}+frac{5}{left(x-3right)left(x-8right)}+frac{12}{left(x-8right)left(x-20right)}-frac{1}{x-20}frac{-3}{4}với xnotin{1;3;8;20}c, TÌM X BIẾT:frac{x-1}{2009}+frac{x-2}{2008}frac{x-3}{2007}+frac{x-4}{2006}GIÚP MÌNH CHÚT NHA MÌNH CẦN NGAY. THANKS!
Đọc tiếp
TÌM x BIẾT:
a,\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
với x\(\notin\){-2;-5;-10;-17}
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
với x\(\notin\){1;3;8;20}
c, TÌM X BIẾT:
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
GIÚP MÌNH CHÚT NHA MÌNH CẦN NGAY. THANKS!
Tim x biet :
\(a,\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(b,\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)
<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
<=> x = 2010
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Bình luận (3)
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)
=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)
=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16
=> -6+16 = 4x - 3x => x = 10
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Bình luận (1)
a)\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{18}\right)x\left(1-\frac{1}{19}\right)x\left(1-\frac{1}{20}\right)\)
b)\(1\frac{1}{2}x1\frac{1}{3}x1\frac{1}{4}x1\frac{1}{5}x......x1\frac{1}{2005}x1\frac{1}{2006}x1\frac{1}{2007}\)
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
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DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
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Tìm x,biết
a, frac{3}{left(x+2right)left(x+5right)}+frac{5}{left(x+5right)left(x+10right)}+frac{7}{left(x+10right)left(x+17right)}frac{x}{left(x+2right)left(x+17right)}
Với x ∉ -2,-5,-10,-17
b,frac{2}{left(x-1right)left(x-3right)}+frac{5}{left(x-3right)left(x-8right)}+frac{12}{left(x-8right)left(x-20right)}-frac{1}{x-20}frac{-3}{4}
Với x∉1,3,8,20
c,frac{x-1}{2009}+frac{x-2}{2008}frac{x-3}{2007}+frac{x-4}{2006}
Đọc tiếp
Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
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Giải phương trình :
a) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) \(6x^4-5x^3-38x^2-5x+6=0\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
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\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
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Forever Miss You nếu (x-2009)=0
thì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\text{ko}?\)
nếu làm cách đó xét 2 trường hợp :")
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