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b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
<=>\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)\( \left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
nên phương trình đó xảy ra khi và chỉ khi x+2009=0
<=>x=-2009
Vậy phương trình có no là x=-2009
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)\)=
\(\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\)\(\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\) \(\left(x+2009\right)\)\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)= 0
\(\Leftrightarrow\)\(x+2009=0\)
( vì \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\) \(\ne0\))
\(\Leftrightarrow\) \(x=-2009\)
Vậy x = -2009
Bạn tham khảo lời giải tại đây:
Câu hỏi của Nguyễn Thị Minh - Toán lớp 8 | Học trực tuyến
ap dung cong thuc: a/b = c/d <=> ad= bc <=> c = ad/b
A = (4x2-7x+3)(x2+2x+1)/(x2-1)
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x+6\right)\left(x-1\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-6^2.P_{min}\Leftrightarrow x^2+5xđạtGTNN\)
\(x^2+5x\ge0\Leftrightarrow x\left(x+5\right)\ge0\)
Dấu "=" xảy ra <=> \(x\in\left\{0;-5\right\}\)
Vậy: Pmin=-36 <=> x E {0;-5}
\(\frac{x}{y^3-1}-\frac{y}{x^3-1}\)
\(=\frac{1-y}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}\)
\(=\frac{-x^2-x-1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)
\(=\frac{\left(y^2-x^2\right)+y-x}{x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1}\)
\(=\frac{\left(y-x\right)\left(y+x\right)+y-x}{x^2y^2+x^2y+xy^2+x^2+xy+y^2+x+y+1}\)
\(=\frac{y-x+y-x}{x^2y^2+xy\left(x+y\right)+x\left(x+y\right)+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+xy+x+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+x\left(y+1\right)+y^2+x+y+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+\left(1-y\right)\left(y+1\right)+y^2+\left(x+y\right)+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+1-y^2+y^2+1+1}\)
\(=\frac{2\left(y-x\right)}{x^2y^2+3}\)
\(\dfrac{1}{{x - 1}} + \dfrac{2}{{x - 2}} + \dfrac{3}{{x - 3}} = \dfrac{6}{{x - 6}}\)
ĐKXĐ: \(x \ne1; x \ne 2;x \ne 3;x \ne6\)
\( \Leftrightarrow \dfrac{{2\left( {x - 3} \right) + 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{{6\left( {x + 1} \right) - \left( {x - 6} \right)}}{{\left( {x - 6} \right)\left( {x - 1} \right)}}\\ \Leftrightarrow \dfrac{{5x - 12}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{{5x}}{{\left( {x - 6} \right)\left( {x - 1} \right)}}\\ \Leftrightarrow \left( {5x - 12} \right)\left( {{x^2} - 7x + 16} \right) = 5x\left( {{x^2} - 5x + 6} \right)\\ \Leftrightarrow - 22{x^2} + 84x - 72 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{21 + 3\sqrt 5 }}{{11}} (tm)\\ x = \dfrac{{21 - 3\sqrt 5 }}{{11}} (tm) \end{array} \right. \)