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theo đề baiif nên
x+1/2008+x+2/2007+x+3/2006-(x+4/2005)-(x+5/2004)-(x+6/2003)=0
suy ra [(x+1/2008)+1]+[(x+2/2007)+1]+[x+3/2006)+1]-[(x+4/2005)+1]-[(x+5/2004)+1]-[(x+6/2003)+1]=0
suy ra (x+2009/2008)+(x+2009/2007)+(x+2009/2006)-(x+2009/2005)-(x+2009/2004)-(x+2009/2003)=0
nên (x+2009)(1/2008+1/2007+1/2006-1/2005-1/2004-1/2003)=0
V1 V2
Dễ thấy V2>0 NÊN x+2009=0 suy ra x=-2009
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Rightarrow\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Rightarrow3x^2-3x=0\)
\(\Rightarrow3x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)
\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)
\(=x-2\sqrt{x}+4\)
=.= hok tốt!!
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
\(\frac{2x}{x^2+4x+4}+\frac{x+1}{x+2}+\frac{2-x}{x^2+4x+4}\)
\(=\frac{2x}{\left(x+2\right)^2}+\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)^2}+\frac{2-x}{\left(x+2\right)^2}\)
\(=\frac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}\)
\(=\frac{x^2+4x+4}{\left(x+2\right)^2}\)
\(=\frac{\left(x+2\right)^2}{\left(x+2\right)^2}\)
\(=1\)
Trả lời:
\(\frac{x+3}{4}=\frac{2-x}{5}\)
\(\Leftrightarrow\frac{5\left(x+3\right)}{20}=\frac{4\left(2-x\right)}{20}\)
\(\Leftrightarrow5\left(x+3\right)=4\left(2-x\right)\)
\(\Leftrightarrow5x+15=8-4x\)
\(\Leftrightarrow5x+4x=8-15\)
\(\Leftrightarrow9x=-7\)
\(\Leftrightarrow x=-\frac{7}{9}\)
Vậy x = - 7/9 là nghiệm của pt.
\(\frac{x+3}{4}=\frac{2-x}{5}\)
\(\Rightarrow\frac{5\left(x+3\right)}{20}=\frac{4\left(2-x\right)}{20}\)
\(\Rightarrow5\left(x+3\right)=4\left(2-x\right)\)
\(\Rightarrow5x+15=8-4x\)
\(\Rightarrow5x+4x=8-15\)
\(\Rightarrow9x=-1\Leftrightarrow x=\frac{-7}{9}\)
Vậy ............