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sua de \(\frac{3}{x^4-x^3+x-1}\) \(-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\) (dk \(x\ne+-1\) )
P=\(\frac{3}{\left(x^2-1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x^2-1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=\(\frac{2}{x^4+x^2+1}>0\)
P\(< \frac{32}{9}\Leftrightarrow\frac{2}{x^4+x^2+1}< \frac{32}{9}\)
\(\Leftrightarrow16x^4+16x^2+7>0\)
\(\Rightarrow\)\(0< P< \frac{32}{9}\) VOI X KHAC 1;-1
\(A=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\)
\(=\frac{3}{\left(x-1\right)\left(x^3+1\right)}-\frac{1}{\left(x+1\right)\left(x^3-1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\left[\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\left[\frac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{x^2-x+1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)\(=\frac{3x^2+3x+3-x^2+x-1}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2-4x-4}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2x^2-2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2}{x^4+x^2+1}\)
\(\Rightarrow A=\frac{2}{x^4+x^2+1}\left(x\ne\pm1\right)\)
Ta có: \(x^4+x^2+1=\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
Vậy A > 0 với mọi \(x\ne\pm1\)
cho hỏi là mẫu biểu thức A là\(\sqrt{x}-3\) hay\(\sqrt{x-3}\)
