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ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)
\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)
\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)
\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)
\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)
\(P=\frac{x^4+1}{2x+1}\)
Vậy \(P=\frac{x^4+1}{2x+1}\)
Bài 1:
\(A=x^2-6x+13=\left(x-3\right)^2+4\ge4\)
Vậy \(Min\)\(A=4\)\(\Leftrightarrow\)\(x=3\)
\(B=2x^2+8x=2\left(x^2+4x+4\right)-8=2\left(x+2\right)^2-8\ge-8\)
Vậy \(Min\)\(B=-8\)\(\Leftrightarrow\)\(x=-2\)
\(C=4x^2+20x=\left(2x+5\right)^2-25\ge-25\)
Vậy \(Min\)\(C=-25\)\(\Leftrightarrow\)\(x=-\frac{5}{2}\)
Bài 3:
a) \(x^2+12x+39=\left(x+6\right)^2+3>0\)
b) \(4x^2+4x+3=\left(2x+1\right)^2+2>0\)
\(P=\dfrac{2x^5-x^4-2x+1}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\dfrac{2x^5-x^4-2x+1}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2}{\left(2x+1\right)}\)
\(P=\dfrac{2x^5-x^4-2x+1+2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(P=\dfrac{2x^5-x^4+2x-1}{\left(2x-1\right)\left(2x+1\right)}\)
\(P=\dfrac{x^4\left(2x-1\right)+2x-1}{\left(2x-1\right)\left(2x+1\right)}\)
\(P=\dfrac{\left(2x-1\right)\left(x^4+1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{x^4+1}{2x+1}\)
cho P=6
\(\dfrac{x^4+1}{2x+1}=6\)
\(\Leftrightarrow x^4+1=6\left(2x+1\right)\)(đk \(x\ne-\dfrac{1}{2}\))
\(\Leftrightarrow x^4-12x-5=0\)
rồi suy ra x
\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)
TKS bạn