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ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)
\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)
\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)
\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)
\(P=\frac{x^4+1}{2x+1}\)
Vậy \(P=\frac{x^4+1}{2x+1}\)
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a: A=(4x+5)^2-2*(4x+5)(4x-5)+(4x-5)^2
=(4x+5-4x+5)^2
=10^2=100
b: B=(3x-2)^2*(3x+2)^2-2(2x+3)(2x-3)
=(9x^2-4)^2-2(4x^2-9)
=81x^4-72x^2+16-8x^2+18
=81x^4-80x^2+34
\(a,A=\left(4x-5\right)^2+\left(4x+5\right)^2+2\left(5+4x\right)\left(5-4x\right)\)
\(=\left(5-4x\right)^2 +2\left(5-4x\right)\left(4x+5\right)+\left(4x+5\right)^2\)
\(=\left(5-4x+4x+5\right)^2\)
\(=10^2\)
\(=100\)
\(b,B=\left(3x-2\right)^2\left(3x+2\right)^2-2\left(2x+3\right)\left(2x-3\right)\)
\(=\left(9x^2-4\right)^2-2\left(4x^2-9\right)\)
\(=81x^4-72x^2+16-8x^2+18\)
\(=81x^4-80x^2+34\)
#\(Urushi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)
\(=8x^3-8x^2+29\)
ta có
\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)
\(A=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)
\(A=8x^3-8x^2+29\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
![](https://rs.olm.vn/images/avt/0.png?1311)